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LIBRARY OF CONGRESS. 

ChV|A.5^!pyriglit Xo.„.._._. 
Shelf. 



UNITED STATES OF AMERICA. 



ELEMENTS 



TEiaONOMETET 



WITH TABLES 



BY 

HERBERT C. WHITAKER, Ph.D. 

CENTRAL MANUAL TRAINING SCHOOL 
PHILADELPHIA, PENNSYLVANIA 



>*X< 



PHILADELPHIA 
D. ANSON PARTRIDGE 

1898 






11017 



Copyright, 1S98, 
By H. C. WHITAKEE. 



iW?,£ co *g; 



JUL -7 1898 



TWO RECEIVED. 

V NortoooU ^ress 

J. S. Cushing & Co. — Berwick & Smith 
Norwood Mass. U.S.A. 



2nd CO 
1898. 



PREFACE. 



The introduction and first five chapters of this book have 
been prepared for the use of beginners in the subject. The 
second appendix has been added for those intending to take 
up work in higher departments of Mathematics. 

The subject of logarithms is usually considered in treatises 
on algebra ; but for the convenience of those not familiar 
with the theory of this subject, the first appendix has been 
added. 

To aid in a clear understanding of each process in the 
various investigations, the aim has been to closely associate 
with every equation a definite meaning with reference to a 
diagram. 

The tables used are in general as precise as the data 
usually obtained by students in engineering, physics, or 
chemistry. It is assumed that a few seven-place tables are 
accessible to a class ; and it is recommended that some class- 
room work be done with seven-place tables, especially in 
the solution of spherical triangles. 

It is hoped that the answers are free from error. The 
numerical computations were made with care by the author, 
but the results have not been verified by any one else. 

Any corrections or suggestions relating to the book will 
be thankfully received. 

H. C. WHITAKER. 

Philadelphia, May, 1898. 



CONTENTS. 



Table of natural values of functions i 

Table of circular measure of angles vii 

Table of hyperbolic functions vii 

INTRODUCTION. 



§1 
§2 
§3 
§4 
§5 
§6 
§7 



Meaning of geometric operations 
Negative segments and angles . 
Products and quotients of segments . 
Homogeneity of trigonometric equations 
Significant figures of numbers . 
Approximate values of numbers 
Contractions in arithmetical operations 



CHAPTER I. 



Functions of Angles. 



§ 8. Generation of an angle 

§ 9. The four quadrants 

§ 10. The six ratios 

§11. Relations between functions and segments 

§ 12. Device for memorizing relations 

§ 13. Absolute values and algebraic signs of functions 

§ 14. Functions of 180° ± x, of 90° + x, and of - x . 

§ 15. The curve of each function .... 

§ 16. Limiting values of functions .... 

§ 17. The number of angles corresponding to a value 

function 

§ 18. Functions of \ x and 2 x . 

§ 19. Functions of particular angles .... 



of a 



11 
12 
14 
17 
19 
20 
22 
24 
26 

26 

28 
29 



CONTENTS. 

CHAPTER II. 
Explanation or the Tables. 

PAGE 

§ 20. The table of natural values of functions . . . .31 

§ 21. Principles of logarithms 37 

§ 22. Use of the table of logarithms 41 

§ 23. The table of logarithms of functions ..... 45 

CHAPTER III. 
Right Triangles. 



§ 24. Problems are indeterminate, ambiguous, determinate, or 

impossible 

§ 25. Relations of the parts of a right triangle 

§ 26. Solutions of triangles using numbers. 

§ 27. Applications 

§ 28. Area of a right triangle 

§ 29. Solutions of triangles using logarithms 

§ 30. Problems 



46 
47 
49 
51 
55 
56 
58 



CHAPTER IV. 

The Isosceles Triangle and the Regular Polygon. 

§ 31. Formulae for isosceles triangles 60 

§ 32. Problems in isosceles triangles ...... 61 

§ 33. Formulae for regular polygons 63 

§ 34. Problems in regular polygons . . . .65 

CHAPTER V. 

Oblique Triangles. 

§ 35. Notation 66 

§ 36. Methods of solution 66 

§ 37. Solutions using numbers 70 

§38. CotB = c cscA -cot A 74 



CONTENTS. 

PAGE 

§ 39. The area of a triangle 76 

§ 40. Solutions using logarithms 77 

§ 41. Applications with solutions 81 

§ 42. Problems 87 



CHAPTER VI. 
Properties of Triangles. 



§ 43. The altitudes, bisectors, and medians 

§ 44. The incircle and the excircles .... 

§ 45. The circumcircle 

§ 46. The sum and difference of two sides and two angles 

§ 47. General method of solution, given three segments 

§ 48. Points of intersection of the circumcircle . 

§ 49. The nine-points circle .... 

§ 50. Abscissae and ordinates of points 

§ 51. Problems 

§ 52. Relations between the angles of a triangle 



CHAPTER VII. 
Spherical Triangles. 

§ 53. The general spherical triangle 

§ 54. Formulae for right triangles 

§ 55. Device for rapidly obtaining formulae 

§ 56. Solution of right triangles 

§ 57. Polar, quadrantal, and isosceles triangles ; the regular 

polygon 

§ 58. Oblique spherical triangles 

§ 59. The incircle and the excircles 

§ 60. The incircle of the polar triangle and the circumcircle 

§ 61. Formulae for oblique triangles ..... 

§ 62. Methods of solution 

§ 63. Solutions 

§ 64. A desired part directly in terms of given parts 

§ 65. Applications to geodesy and astronomy . 

§ QQ. Area of a spherical triangle ..... 



94 

96 
98 
98 
100 
100 
102 
104 
105 
108 



110 
112 
114 
115 

117 
120 
120 
121 
122 
122 
123 
126 
128 
132 



CONTENTS. 

APPENDIX L 

Theory of Logarithms. 

§ 67. Arithmetic and geometric series 

§ 68. Napierian logarithms treated geometrically 

§ 69. Present systems treated geometrically 

§ 70. Natural logarithms 

§ 71. The number corresponding to a given logarithm 

§ 72. A logarithm is an exponent of a base 

§ 73. The logarithm corresponding to a given number 

§ 74. The logarithmic curve 

§ 75. Problems 



PAGE 

133 

134 
136 
137 
138 
140 
141 
143 
144 



APPENDIX II. 

Goniometry, Complex Quantities, Hyperbolic Functions. 

§ 76. Functions of the sum or difference of angles . . . 144 

§ 77. Trigonometric equations ; the quadratic equation . . 146 

§ 78. The circular measure of an angle . - . . . . 149 

§ 79. Arc functions 150 

§ 80. Complex Quantities : addition, subtraction, multiplication, 

division, powers and roots ...... 152 

§ 81. Functions of multiple angles 159 

§ 82. Exponential values of sin 6 and cos d 159 

§83. Hyperbolic Functions: the equilateral hyperbola . . 160 

§ 84. Definitions of hyperbolic functions 161 

§ 85. Exponential values of hyperbolic functions . . . 162 

§ 86. General relations 163 

§ 87. Longitude of it 164 

§ 88. Solution of the cubic equation 164 

Problems eor Examination 167 

Answers to Problems . . . . . . . .177 

Table or Logarithms of Numbers ...... viii 

Table of Logarithms of Functions ..... x 

Table of Formulae xvi 

Table of Constants xvi 



A TABLE 

OF THE 

NATURAL VALUES OF THE TRIGONOMETRIC 
FUNCTIONS OF ANGLES 

FROM 0° TO 90° FOR EVERY 10' 



A TABLE 

FOR CONVERTING THE SEXAGESIMAL MEASURE 

OF AN ANGLE INTO CIRCULAR MEASURE 



A TABLE 

FOR CONVERTING CIRCULAR FUNCTIONS INTO 

HYPERBOLIC FUNCTIONS 



NATURAL VALUES OF FUNCTIONS. 



X 


sin* 


tan* 


seer 


CSC* 


cot* 


cos* 







.0 00000 


000000 


1.00 00 


Infinite. ] 


infinite. 


1.00 00 


90 


10' 


02909 


02909 


00 


343.775 


343.774 


00 


50' 


20' 


05818 


05S1S 


00 


171.SS8 


171.885 


00 


40' 


30' 


08727 


0S727 


00 


114.593 


114.5S9 


00 


30' 


40' 


11635 


11636 


01 


85.946 


S5.940 


.99 99 


20' 


50' 


14544 


14545 


01 


6S.757 


6S.750 


99 


10' 


1° 


.0 1745 


.0 1746 


1.00 02 


57.299 


57.290 


.99 98 


89 


10' 


2036 


2036 


02 


49.114 


49.104 


9S 


50' 


20' 


2327 


2328 


03 


42.976 


42.964 


97 


40' 


30' 


261S 


2619 


03 


38.202 


3S.188 


97 


30' 


40' 


290S 


2910 


04 


34.382 


34.36S 


96 


20' 


£<y 


3199 


3201 


05 


31.258 


31.242 


95 


10' 


2 


.0 3490 


.0 3492 


1.00 06 


28.654 


2S.636 


.99 94 


88 3 


10' 


3781 


3783 


07 


26.451 


26.432 


93 


50' 


20' 


4071 


4075 


08 


24.562 


24.542 


92 


40' 


30' 


4362 


4366 


10 


22.926 


22.904 


90 


30' 


40' 


4653 


4658 


11 


21.494 


21.470 


89 


20' 


50' 


4943 


4949 


12 


20.230 


20.206 


SS 


10' 


3" 


.0 5234 


.0 5241 


1.0014 


19.107 


19.081 


.99 86 


87 : 


10' 


5524 


5533 


15 


18.103 


1S.075 


85 


50' 


20' 


5S14 


5S24 


17 


17.195 


17.169 


83 


40' 


30' 


6105 


6116 


19 


16.3S0 


16.350 


81 


30' 


40' 


6395 


640S 


21 


15.637 


15.605 


80 


20' 


50' 


66S5 


6700 


22 


14.958 


14.924 


78 


10' 


4 : 


.0 6976 


.0 6993 


1.00 24 


14.336 


14.301 


.99 76 


86 : 


10' 


7266 


7235 


26 


13.763 


13.727 


74 


50' 


20' 


7556 


7578 


29 


13.235 


13.197 


71 


40' 


30' 


7S46 


7S70 


31 


12.745 


12.706 


69 


30' 


40' 


8136 


8163 


33 


12.291 


12.250 


67 


20' 


50' 


8426 


8456 


36 


11.86S 


11.826 


64 


10' 


5 


.0 8716 


.0 S749 


1.00 3S 


11.474 


11.430 


.99 62 


85 : 


10' 


9005 


9042 


41 


11.105 


11.059 


59 


50' 


20' 


9295 


9335 


43 


10.758 


10.712 


57 


40' 


30' 


9585 


9629 


46 


10.433 


10.385 


54 


30' 


40' 


9S74 


9923 


49 


10.128 


10.078 


51 


20' 


50' 


.1 0164 


.10216 


52 


9.8391 


9.7882 


48 


10' 


6 Z 


.1 0453 


.10510 


1.00 55 


9.566S 


9.5144 


.99 45 


84 : 


10' 


0742 


0805 


55 


9.3092 


9.2553 


42 


50' 


20' 


1031 


1099 


61 


9.0652 


9.0098 


39 


40' 


30' 


1320 


1394 


65 


8.8337 


8.7769 


36 


30' 


40' 


1609 


16SS 


6S 


8.6138 


8.5555 


32 


20' 


50' 


1S9S 


1983 


72 


8.4047 


8.3450 


29 


10' 


V 


.12157 


.12275 


1.00 75 


8.2055 


S.1443 


.99 25 


83 


1 


cos/ 


cot/ 


CSC/ 


sec/ 


tan/ 


sin/ 


/ 



NATURAL VALUES OF FUNCTIONS. 



X 


sin* 


tan* 


sec* 


CSC* 


cot* 


cos* 




7° 


.1 219 


.1 228 


1.00 75 


8. 2055 


8. 1443 


.99 25 


83° 


10' 


248 


257 


79 


0156 


7. 9530 


22 


50' 


20' 


276 


287 


82 


7. 8344 


7704 


18 


40' 


30' 


305 


317 


86 


6613 


5958 


14 


30' 


40' 


334 


346 


90 


4957 


4287 


11 


20' 


50' 


363 


376 


94 


3372 


2687 


07 


10' 


8° 


.1392 


.1405 


1.00 98 


7. 1853 


7. 1154 


.99 03 


82° 


10' 


421 


435 


1.01 02 


0396 


6. 9682 


.98 99 


50' 


20' 


449 


465 


07 


6. 8998 


8269 


94 


40' 


30' 


478 


495 


11 


7655 


6912 


90 


30' 


40' 


507 


524 


15 


6363 


5606 


86 


20' 


50' 


536 


554 


20 


5121 


4348 


81 


10' 


9° 


.1564 


.1584 


1.01 25 


6. 3925 


6. 3138 


.98 77 


81° 


10' 


593 


614 


29 


2772 


1970 


72 


50' 


20' 


622 


644 


34 


1661 


0844 


68 


40' 


30' 


650 


673 


39 


0589 


5. 9758 


63 


30' 


40' 


679 


703 


44 


5. 9554 


8708 


58 


20' 


50' 


708 


733 


49 


8554 


7694 


53 


10' 


10° 


.1 736 


.1763 


1.01 54 


5. 7588 


5. 6713 


.98 48 


80° 


10' 


765 


793 


60 


6653 


5764 


43 


50' 


20' 


794 


823 


65 


5749 


4845 


38 


40' 


30' 


822 


853 


70 


4874 


3955 


33 


30' 


40' 


851 


883 


76 


4026 


3093 


27 


20' 


50' 


880 


914 


81 


3205 


2257 


22 


10' 


11° 


.1908 


.1944 


1.01 87 


5. 2408 


5. 1446 


.98 16 


79° 


10' 


937 


974 


93 


1636 


0658 


11 


50' 


20' 


965 


.2 004 


99 


0886 


4. 9894 


05 


40' 


30' 


994 


035 


1.02 05 


0159 


9152 


.97 99 


30' 


40' 


.2 022 


065 


11 


4. 9452 


8430 


93 


20' 


50' 


051 


095 


17 


8765 


7729 


87 


10' 


12° 


.2 079 


.2 126 


1.02 23 


4. 8097 


4. 7046 


.97 81 


78° 


10' 


108 


156 


30 


7448 


6382 


75 


50' 


20' 


136 


186 


36 


6817 


5736 


69 


40' 


30' 


164 


217 


43 


6202 


5107 


63 


30' 


40' 


193 


247 


49 


5604 


4494 


57 


20' 


50' 


221 


278 


56 


5022 


3897 


50 


10' 


13° 


.2 250 


.2 309 


1.02 63 


4. 4454 


4. 3315 


.97 44 


77° 


10' 


278 


339 


70 


3901 


2747 


37 


50' 


20' 


306 


370 


77 


3362 


2193 


30 


40' 


30' 


334 


401 


84 


2837 


1653 


24 


30' 


40' 


363 


432 


91 


2324 


1126 


17 


20' 


50' 


391 


462 


99 


1824 


0611 


10 


10' 


14° 


.2 419 


.2 493 


1.03 06 


4. 1336 


4. 0108 


.97 03 


76° 




cos/ 


cot/ 


CSC/ 


sec/ 


tan/ 


sin/ 


/ 



Ill 



NATURAL VALUES OF FUNCTIONS. 



* 


sin* 


tan* 


sec* 


CSC* 


cot* 


cos* 




14 : 


.2 419 


.2 493 


1.03 06 


4. 1336 


4. 010S 


.97 03 


76 : 


10' 


447 


524 


14 


0S59 


3. 9617 


.96 96 


50' 


20' 


476 


555 


21 


0394 


9136 


89 


40' 


30' 


504 


556 


29 


3. 9939 


8667 


SI 


30 f 


40' 


532 


617 


37 


9495 


5205 


74 


20' 


50' 


560 


64S 


45 


9061 


7760 


67 


10' 


15 : 


.2 5S8 


.2 679 


1.03 53 


3. S637 


3. 7321 


.96 59 


75 : 


10' 


616 


711 


61 


S222 


6S91 


52 


50' 


20' 


644 


742 


69 


7517 


6470 


44 


40' 


30' 


672 


773 


77 


7420 


6059 


36 


30' 


40' 


700 


805 


S6 


7032 


5656 


28 


20' 


50' 


728 


S36 


94 


6652 


5261 


21 


10' 


16= 


.2 756 


.2 867 


1.04 03 


3. 6280 


3. 4S74 


.96 13 


74 : 


10' 


754 


S99 


12 


5915 


4495 


05 


50' 


20' 


812 


931 


21 


5559 


4124 


.95 96 


40' 


30' 


840 


962 


29 


5209 


3759 


SS 


30' 


40' 


S6S 


994 


39 


4567 


3402 


SO 


20' 


50' 


S96 


.3 026 


4S 


4532 


3052 


72 


10' 


17 : 


.2 924 


.3 057 


1.04 57 


3. 4203 


3.2709 


.95 63 


73 : 


10' 


952 


0S9 


66 


3SS1 


2371 


55 


50' 


20' 


979 


121 


76 


3565 


2041 


46 


40' 


30' 


.3 007 


153 


85 


3255 


1716 


37 


30' 


40' 


035 


185 


95 


2951 


1397 


28 


20' 


50' 


062 


217 


1.05 05 


2653 


10S4 


20 


10' 


18 : 


.3 090 


.3 249 


1.05 15 


3. 2361 


3. 0777 


.95 11 


72 : 


10' 


US 


231 


25 


2074 


0475 


02 


50' 


20' 


145 


314 


35 


1792 


0178 


.94 92 


40' 


30' 


173 


346 


45 


1515 


2. 95S7 


S3 


30' 


40' 


201 


37S 


55 


1244 


9600 


74 


20' 


50' 


228 


411 


66 


0977 


9319 


65 


10' 


19 : 


.3 256 


.3 443 


1.05 76 


3. 0716 


2.9042 


.94 55 


71° 


10' 


283 


476 


S7 


0458 


S770 


46 


50' 


20' 


311 


508 


9S 


0206 


8502 


36 


■40' 


30' 


33S 


541 


1.06 OS 


2.9957 


S239 


26 


30' 


40' 


365 


5 74 


19 


9713 


79S0 


17 


20' 


50' 


393 


607 


31 


9474 


7725 


07 


10' 


20 : 


.3 420 


.3 640 


1.06 42 


2. 923S 


2. 7475 


.93 97 


70 : 


10' 


44S 


673 


53 


9006 


722S 


87 


50' 


20' 


475 


706 


65 


S779 


6985 


77 


40' 


30' 


502 


739 


76 


S555 


6746 


67 


30' 


40' 


529 


772 


8S 


S334 


6511 


56 


20' 


50' 


55 7 


805 


1.07 00 


8117 


6279 


46 


10' 


21° 


.3 584 


.3S39 


1 07 11 


2. 7904 


2. 6051 


.93 36 


69 : 




cos/ 


cot/ 


CSC/ 


sec/ 


tan/ 


sin/ 


/ 



NATURAL VALUES OF FUNCTIONS. 



IY 



X 


sin* 


tan* 


sec* 


CSC* 


cot* 


cos* 




21° 


.3 584 


.3 839 


1.07 11 


2. 7904 


2. 6051 


.93 36 


69° 


10' 


611 


872 


23 


7695 


5826 


25 


50' 


20' 


638 


906 


36 


7488 


5605 


15 


40' 


30' 


665 


939 


48 


7285 


5386 


04 


30' 


40' 


692 


973 


60 


7085 


5172 


.92 93 


20' 


50' 


719 


.4 006 


73 


6888 


4960 


83 


10' 


22° 


.3 746 


.4 040 


1.07 85 


2. 6695 


2. 4751 


.92 72 


68° 


10' 


773 


074 


98 


6504 


4545 


61 


50' 


20' 


800 


108 


1.08 11 


6316 


4342 


50 


40' 


30' 


827 


142 


24 


6131 


4142 


39 


30' 


40' 


854 


176 


37 


5949 


3945 


28 


20' 


50' 


881 


210 


50 


5770 


3750 


16 


10' 


23° 


.3 907 


.4 245 


1.08 64 


2. 5593 


2. 3559 


.92 05 


67° 


10' 


934 


279 


77 


5419 


3369 


.9194 


50' 


20' 


961 


314 


91 


5247 


3183 


82 


40' 


30' 


987 


348 


1.09 04 


5078 


2998 


71 


30' 


40' 


.4 014 


383 


18 


4912 


2817 


59 


20' 


50' 


041 


417 


32 


4748 


2637 


47 


10' 


24° 


.4 067 


.4 452 


1.09 46 


2. 4586 


2. 2460 


.9135 


66° 


10' 


094 


487 


61 


4426 


2286 


24 


50' 


20' 


120 


522 


75 


4269 


2113 


12 


40' 


30' 


147 


557 


89 


4114 


1943 


00 


30' 


40' 


173 


592 


1.10 04 


3961 


1775 


.90 88 


20' 


50' 


200 


628 


19 


3811 


1609 


75 


10' 


25° 


.4 226 


.4 663 


1.10 34 


2. 3662 


2. 1445 


.90 63 


65° 


10' 


• 253 


699 


49 


3515 


1283 


51 


50' 


20' 


279 


734 


64 


3371 


1123 


38 


40' 


30' 


305 


770 


79 


3228 


0965 


26 


30' 


40' 


331 


806 


95 


3087 


0809 


13 


20' 


50' 


358 


841 


1.11 10 


2949 


0655 


01 


10' 


26° 


.4 384 


.4 877 


1.11 26 


2. 2812 


2. 0503 


.89 88 


64° 


10' 


410 


913 


42 


2677 


0353 


75 


50' 


20' 


436 


950 


58 


2543 


0204 


62 


40' 


30' 


462 


986 


74 


2412 


0057 


49 


30' 


40' 


488 


.5 022 


90 


2282 


1. 9912 


36 


20' 


50' 


514 


059 


1.12 07 


2153 


9768 


23 


10' 


27° 


.4 540 


.5 095 


1.12 23 


2. 2027 


1. 9626 


.89 10 


63° 


10' 


566 


132 


40 


1902 


9486 


.88 97 


50' 


20' 


592 


169 


57 


1779 


9347 


84 


40' 


30' 


617 


206 


74 


1657 


9210 


70 


30' 


40' 


643 


243 


91 


1537 


9074 


57 


20' 


50' 


669 


280 


1.13 08 


1418 


8940 


43 


10' 


28° 


.4 695 


.5 317 


1.13 26 


2. 1301 


1. 8807 


.88 29 


62° 




cos/ 


cot/ 


CSC/ 


sec/ 


tan/ 


sin/ 


/ 



NATURAL VALUES OF FUNCTIONS. 



X 


sin* 


tan* 


seer 


cscjt 


cot* 


cos* 




28 c 


.4 695 


.5 317 


1.1 326 


2.1 301 


1.8 S07 


.SS29 


62 : 


10' 


720 


354 


343 


185 


676 


16 


50' 


20' 


746 


392 


361 


070 


546 


02 


40' 


30' 


772 


430 


379 


2.0 957 


41S 


.S7 8S 


30' 


W 


797 


467 


397 


S46 


291 


74 


20' 


50' 


S23 


505 


415 


736 


165 


60 


10' 


29 : 


.4S4S 


.5 543 


1.1 434 


2.0 627 


1.8 040 


.S7 46 


61 3 


10' 


S74 


5S1 


452 


519 


1.7 917 


32 


50' 


20' 


899 


619 


471 


413 


796 


IS 


40' 


30' 


924 


65 S 


490 


308 


675 


04 


30' 


40' 


950 


696 


509 


204 


556 


.S6S9 


20' 


50' 


975 


735 


52S 


101 


437 


75 


10' 


30 : 


.5 000 


.5 774 


1.1 547 


2.0 000 


1.7 321 


.S6 60 


60° 


10' 


025 


812 


566 


1.9 900 


205 


46 


50' 


20' 


050 


S51 


5S6 


801 


090 


31 


40' 


30' 


075 


890 


606 


703 


1.6 977 


16 


30' 


40' 


100 


930 


626 


606 


S64 


01 


20' 


50' 


125 


969 


646 


511 


753 


.s : s; 


10' 


31 : 


.5 150 


.6 009 


1.1 666 


1.9 416 


1.6 643 


.85 72 


59° 


10' 


175 


04S 


6S7 


323 


534 


57 


50' 


20' 


200 


088 


707 


230 


426 


42 


40' 


30' 


225 


128 


725 


139 


319 


26 


30' 


40' 


250 


168 


749 


04S 


212 


11 


20' 


50' 


275 


208 


770 


1.8 959 


107 


.S4 96 


10' 


32° 


.5 299 


.6 249 


1.1 792 


1.8 S71 


1.6 003 


.S4S0 


58° 


10' 


324 


2S9 


813 


783 


1.5 900 


65 


50' 


20' 


34S 


330 


S35 


697 


79S 


50 


40' 


30' 


373 


371 


S57 


612 


697 


34 


30' 


40' 


39S 


412 


S79 


527 


597 


IS 


20' 


50' 


422 


453 


901 


443 


497 


03 


10' 


33 : 


.5 446 


.6 494 


1.1 924 


1.8 361 


1.5 399 


.S3S7 


57° 


10' 


471 


536 


946 


279 


301 


71 


50' 


20' 


495 


577 


969 


19S 


204 


55 


40' 


30' 


519 


619 


992 


118 


10S 


39 


30' 


40' 


544 


661 


1.2 015 


039 


013 


23 


20' 


50' 


56S 


703 


039 


1.7 960 


1.4 919 


07 


10' 


34 : 


.5 592 


.6 745 


1.2 062 


1.7 SS3 


1.4 S26 


.S2 90 


56 


10' 


616 


7S7 


086 


S06 


733 


74 


50' 


20' 


640 


830 


110 


730 


641 


58 


40' 


30' 


664 


S73 


134 


655 


550 


41 


30' 


40' 


6SS 


916 


15S 


581 


460 


25 


20' 


50' 


712 


959 


1S3 


507 


370 


OS 


10' 


35 


.5 736 


.7 002 


1.2 20S 


1.7 434 


1.4 2S1 


.8192 


55° 




cos/ 


cot/ 


esc/ 


sec/ 


tan/ 


sin/ 


/ 



NATURAL VALUES OF FUNCTIONS. 



VI 



* 


sin* 


tan* 


sec* 


CSC* 


cot* 


cos* 




35° 


.5 736 


.7 002 


1.2 208 


1.7 434 


1.4 281 


.8192 


55° 


10' 


760 


046 


233 


362 


193 


75 


50' 


20' 


783 


089 


258 


291 


106 


58 


40' 


30' 


807 


133 


283 


221 


019 


41 


30' 


40' 


831 


177 


309 


151 


1.3 934 


24 


20' 


50' 


854 


221 


335 


081 


848 


07 


10' 


36° 


.5 878 


.7 265 


1.2 361 


1.7 013 


1.3 764 


.80 90 


54° 


10' 


901 


310 


387 


1.6 945 


680 


73 


50' 


20' 


925 


355 


413 


878 


597 


56 


40' 


30' 


948 


400 


440 


812 


514 


39 


30' 


40' 


972 


445 


467 


746 


432 


21 


20' 


50' 


995 


490 


494 


681 


351 


04 


10' 


37° 


.6 018 


.7 536 


1.2 521 


1.6 616 


1.3 270 


.79 86 


53° 


10' 


041 


581 


549 


553 


190 


69 


50' 


20' 


065 


627 


577 


489 


111 


51 


40' 


30' 


088 


673 


605 


427 


032 


34 


30' 


40' 


111 


720 


633 


365 


1.2 954 


16 


20' 


50' 


134 


766 


661 


303 


876 


.78 98 


10' 


38° 


.6 157 


.7 813 


1.2 690 


1.6 243 


1.2 799 


.78 80 


52° 


10' 


180 


860 


719 


183 


723 


62 


50' 


20' 


202 


907 


748 


123 


647 


44 


40' 


30' 


225 


954 


778 


064 


572 


26 


30' 


40' 


248 


.8 002 


807 


005 


497 


08 


20' 


50' 


271 


050 


837 


1.5 948 


423 


.77 90 


10' 


39° 


.6 293 


.8 098 


1.2 868 


1.5 890 


1.2 349 


.77 71 


51° 


10' 


316 


146 


898 


833 


276 


53 


50' 


20' 


338 


195 


929 


777 


203 


35 


40' 


30' 


361 


243 


960 


721 


131 


16 


30' 


40' 


383 


292 


991 


666 


059 


.76 98 


20' 


50' 


406 


342 


1.3 022 


611 


1.1 988 


79 


10' 


40° 


.6 428 


.8 391 


1.3 054 


1.5 557 


1.1918 


.76 60 


50° 


10' 


450 


441 


086 


504 


847 


42 


50' 


20' 


472 


491 


118 


450 


778 


23 


40' 


30' 


494 


541 


151 


398 


708 


04 


30' 


40' 


517 


591 


184 


345 


640 


.75 85 


20' 


50' 


539 


642 


217 


294 


571 


66 


10' 


41° 


.6 561 


.8 693 


1.3 250 


1.5 243 


1.1 504 


.75 47 


49° 


10' 


583 


744 


284 


192 


436 


28 


50' 


20' 


604 


796 


318 


141 


369 


09 


40' 


30' 


626 


847 


352 


092 


303 


.74 90 


30' 


40' 


648 


899 


386 


042 


237 


70 


20' 


50' 


670 


952 


421 


1.4 993 


171 


51 


10' 


42° 


.6 691 


.9 004 


1.3 456 


1.4 945 


1.1 106 


.74 31 


48° 




cos/ 


cot/ 


CSC/ 


sec/ 


tan/ 


sin/ 


/ 



Til 



NATURAL VALUE.' 



>F FUNCTIONS. 



X 


sin jt 


tan i 


sec jr 


csc jt 


cot X 


COS X 




42 : 


.<: :>1 


.:- :■:- 


13 456 


1.4 945 


:.: :>: 


.74 31 


48 : 


10 1 


713 


: :_ 


-:-2 


897 


041 


12 


50* 


2xy 


- - 


110 


: :7 


849 


1.0 977 


\ - 


-: 


3c 


756 


163 


563 


V.I 


913 


73 


30* 


40* 


777 


217 


-:•>: 


755 


550 


: 3 


::■ 


5 ; 


799 


271 


636 


709 


786 


:: 


10* 


43 : 


.6S20 


.9 325 


73 


1.4 663 


: : -:- 


.73 14 


47 3 


io* 


S41 


3S0 


711 


617 


661 


.72 94 


50 1 


yy 


Sf2 


«5 


74S 


572 


599 


74 


40* 


3(y 


55- 


490 


7-: 


: :7 


555 


54 


30* 


40 1 


905 


545 


S24 


-53 


--; 


34 


:; 


5W 


::: 


601 


S-f 3 


439 


416 


14 


1(K 


44 : 


.6 947 


..- -: : 7 


1.5 ;-:2 


1.4 396 


: : ; :: 


.71 93 


46 : 


10 1 


H7 


713 


941 


352 


295 


73 


-:■ 


2C' 


:-ss 


--. 


rs: 


310 


235 


53 


-: 


3(y 


.7 009 


527 


1.4 020 


267 


176 


33 


30* 


4G' 


030 


554 


:-:: 


225 


117 


12 


::■ 


so 1 


050 


942 


101 


153 


05S 


.70 92 


10* 


45 : 


.7 071 1.0 000 


1.4 142 


1.4 142 


1.0 000 


.:: :: 


45 : 




cos/ 


cot/ 


CSC/ 


sec/ 


tan/ 


sin/ 


/ 



" : b : teas 
-Measube. 



X 

ISO 1 



Hypebbolic '' sinh u 
Fr>XTioys. '-coshi; 



tan jr. 

sec r. 



* 


e 


X 


e 


X 


u 


X 


u 


1 ; 


.017- : 


V 


.33:23 


i- 


'jI-'j 


-: 


-:--_, 


?- 


.03491 


: 


.::: r 5 


4= 


.V.Yi 


4.? 


'- .S092 


':- 


.05236 


3 


,::-;37 


5° 


.1049 


— 


3 ;.";: 


Y- 


.06981 


¥ 


.■:■:::-: 


3 : 


.1401 


-- 


: .9063 


5 : 


:S727 


5 


.00145 


10° 


.1754 


-5 


: .9575 


& 


.10472 


6' 


.00175 


12 : 


.2110 


r " 


; . _ 


r- 


.12217 


V 


.;■:::- 


14 ; 


.2-5 


: : 


- - ,- : " 


8° 


.13963 


3 


.00233 


16° 


,2^5J 


;_1 


: _ _ ~-L7 


gp 


.15708 


9' 


.00262 


:s : 


3195 




: 3.3553 


li 




.17453 


10* 


.012?: 


iv- 


.? : -:- 


55 


: 1.2492 


- 




34907 


20* 


.:-: r -2 


11- 


.3;-:S 


60 


: 13170 






: :h; 


30' 


.0:573 


1-- 


.4317 


62 


z 13S90 


- 




. •: r : - :■ 


40 1 


.01164 


2& 


.4702 


64 


: 1.4659 


50' 


£7266 


50* 


.01454 


23 : 


-'- 


66 


B I."- - 


60° 


1.0471 


60* 


.01745 


30 3 


.5493 


65 


D 1.6379 


:: 


1.22173 






: : 


.-?:•:• 


70 


1.7354 


5; : 


139-1: 


10" 


.00005 


:- : 


.6317 


72 


: 1.5-: - 


:. " : 


1.57353 


:: 


.:•:•:•:■: 




.6743 


74 


1.9623 


:;•:- 


1.7-5: : 


40"' 


::■::;• 


- s - 


.71S0 


7-: 


2J0973 



ELEMENTS OF TRIGONOMETRY. 



>^c 



INTRODUCTION. 

Trigonometry investigates the numerical relations be- 
tween segments,* angles, and areas of figures. 

1. The meaning of geometric operations will be retained in 
trigonometry. Thus, Fig. 1, if the segment CD, of length 
a, is added to the segment DE, 
of length b, the sum is the seg- 
ment CE, of length a + b. 

In Fig. 2, if DE, of length b, 
is subtracted from CD, of length 
a, the remainder CE is of length 
a — b. 

In Fig. 3, if CD is of length a, FlG 3> 

CE = a-\-a-\-a-\-a = 4: a. 

In Fig. 4, if CD is of length a, CE (two of the ten equal 
parts of CD) is of length .2 a. 

Hence if a segment , „ 

I -i 1 1 1 " 1 1 1 > 1 

of length a is mul- C E E> 

tiplied by 2.4, the FlG - 4 - 

product will be considered either a divided by 10 and the 

* In this work a straight line, or simply a line, is conceived to be 
infinitely long in two opposite directions from a point ; hence a half- 
line is conceived to be infinitely long in one direction from a point ; 
and a segment of a line, or simply a segment, is the name which will 
be applied to a limited portion of a line. 

1 







a , 


b 




c 




D 
Fig. 1. 

a 




E 


c 




E b 
Fig. 2. 




D 




a 


a a 


a 





INTRODUCTION. 



quotient multiplied by 24, or a multiplied by 24 and 
the product divided by 10. 

In Fig. 5, EF, of length b, may be conceived as being 
divided into CD, of length a. If CD contains EF five times, 



the ratio of CD to E F 



EF is said to be 5. h 
If it is desired to . 



b 



obtain the ratio of c G D 

Fir 5 

EF to CD, CD is 

first divided into 10 equal parts and one of these parts 
CG applied to EF. If EF contains CG exactly two times, 

the ratio of EF to CD is said to be .2. The ratios - = 5 

b b 

and -= .2 have a product of 1, and are said to be reciprocal 
a 

ratios. 

In Fig. 6, EF is not contained exactly in CD, but is con- 
tained twice and a remainder GD more than three-tenths of 

EF, but less than three , , 

and one-half tenths ; the ^ * _JL 

ratio of CD to EF is C~~ G ' ' D 

said to be 2.3 + . This FlG - 6 - 

value of the ratio contains an error not greater than .05. 
The ratio of the error to the assumed value is not greater 
than .05 in 2.3, or 5 in 230, or 1 in 46. If this error in 2.3, 
the assumed value of the ratio, is too great for use in a 
problem in which the ratio enters, EF is divided into 100, 
1000, 10,000, etc., equal parts, and one of them applied to 
GD. Thus, if EF is divided into 10,000 parts and GD is 
found to contain one of these parts not less than 3470.5 
times, but less than 3471 times, the ratio of CD to EF is 
said to be 2.3471 -. This value of the ratio contains an 
error not greater than .00005. The ratio of the error to the 
assumed value is not greater than .00005 in 2.3471, or 1 in 
46,942. 

In the same way, if the ratio of CD to EF is desired, CD 




NEGATIVE SEGMENTS AND ANGLES. g 

is divided into 10, 100, 1000, 10,000, etc., equal parts, and 
so divided applied to EF. Suppose CD is divided into 
10,000 parts, and the beginning of CD is placed at E. 
Then if the 4261st division of CD is the one nearest to F, 
the ratio of EF to CD is said to be approximately .4261 ; 
the ratio of the error to the assumed value of the ratio in 
this case is not greater than 1 in 8522. 

Operations upon angles. — All that has been said with 
reference to addition, subtraction, multi- 
plication, and division of segments will 
apply to corresponding operations per- 
formed upon angles. Thus, Fig. 7, the 
sum of angle BAC(=a) and angle 
CAD (= b) is the angle BAD(= a + b). Fig. 

Similarly, the difference of BAC( = a) 
and CAD(=b), Fig. 8, is the angle 
BAD{=a-b). 

In a similar manner, the interpreta- 
tion of the multiplication of an angle by 
a number, the division of an angle by a 
number, and the division of an angle by 
another angle will be conceived to be the same as in 
geometry. 

NEGATIVE SEGMENTS AND ANGLES. 

2. Assume the direction of measurement of the segment 
CD, of length a, Fig. 9, to be from b 

left to right. From D, the ter- E C a D 

minal point of CD, lay off a seg- FlG - 9 * 

ment DE, of length b, backward on CD. Then if b is 
greater than a, E will be to the left of C, but a — b will 
still be said to be the length of the segment CE, measured 
from C to E, or in the direction opposite to that assumed 
to be positive. Now the length of a in the discussion is 




4 INTRODUCTION. 

arbitrary, and may be conceived to be eqnal to the length 
of CD in Fig. 10, or zero ; in which case the segment CE 

(measured from right to left, or | b p 

opposite the direction assumed to E C 

be positive) is equal to — b. FlG " 10, 

Hence, if any segment is multiplied by — 1, the effect of 
the operation is to measure the segment from the same point 
as at first but in an opposite direction ; thus, — 1 x a — — a, 
and — 1 x — a = + a. 

The multiplication of a segment by a positive number will 
be considered to be repetition of the segment in the same 
direction the number of times indicated by the multi- 
plier (which may be either integral or fractional) ; thus, 
4.3 times — a = — (4.3 a). The product of a segment by a 
negative number will be conceived to be : first, a reversal 
of the direction of the segment, and then the multiplication 
by a positive number ; thus, 

— 4.3 x a = 4.3 x — 1 x a = 4.3 x — a = — (4.3 a). 
Also 

- 4.3 x - a = 4.3 x - 1 x - a = 4.3 x a = + (4.3 a). 

If division is defined as being the process of obtaining a 
quantity, which, multiplied with the divisor, will produce 
the dividend, each process of multiplication gives 

-(43a) = _ a 
4.3 

-(4.3 a) 

-=Ts = a - 

+ (4-3 a) _ . 

~T43 

The same methods of reasoning will give the conception 
of negative angles. Thus, if turning a screw in one direction 
is positive turning, then turning the screw in the opposite 
direction is negative turning, because it cancels or undoes 



-(4.3a)_ 


4.3. 


— a 


-(4.3 a) _ 


-4.3. 


a 


+ (4.3 a) _ 


43 


— a 





PKODUCTS AND QUOTIENTS OF SEGMENTS. 5 

an equal amount of positive turning ; and negative angles so 
conceived may be multiplied by positive or negative num- 
bers, or divided by positive or negative numbers or angles. 



PKODUCTS AND QUOTIENTS OF SEGMENTS. 

3. In geometry, the product of two segments is explained 
to be a term used to aid us in counting the number of units 
of area in a surface ; thus if a rectangle has sides of length 
3 inches and 2 inches, respectively, the area is said to be 
3 inches x 2 inches = 6 square inches. This statement is 
a short way of saying that in the rectangle there can be 
arranged 2 rows of inch-squares, 3 in a row, making a 
total of 6 inch-squares or 6 square inches. In the same 
way the operation 4 inches x 3 inches x 2 inches = 24 cubic 
inches, is a method of counting cubes in a rectangular par- 
allelopiped. In trigonometry, the same meaning is attached 
to these expressions as in geometry. Thus, the product of 
a segment and an abstract number, say 4.3 a, is said to be a 
length; the product (so called) of two segments and an 
abstract number, say 4.3 a 2 or 4.3 ab, is said to be an area ; 
the product of three segments and an abstract number, say 
4.3 abc or 4.3 a 2 b, is said to be a volume. 

Since division is the inverse of multiplication, an area is 

indicated by the quotient of a volume by a segment, such 

4.3 abc a . .-. t 4.3 abc 4.3 ab , . , 3rm — r- 

aS HT Simdarly ' 5^fe' 5^7' V43a *> V4 ' 3a&0 

represent segments. 

Expressions having no geometric meaning. — In geometry, 
expressions for lengths, areas, and volumes have meaning; 
but many forms which occur in trigonometric reductions 

such as Va, - 1 -, Vafrc, 4.3 abed, are said to have no geo- 
a 

metric meaning. 



6 INTRODUCTION. 

Dimensions. — Any expression representing a length is 
said to be of one dimension: an expression representing an 
area is said to be of two dimensions ; and any expression 
is of Ji dimensions if the algebraic sum of the exponents of 
the segments in each term is ft. Thus a ratio has zero 

a ° 

dimensions, since it is of the form - = oft -1 : - has — 1 

3 11 b ° 

dimensions: a ab = a*b 3 is of -| dimensions: abed is of 4 
dimensions : and so on. 

4. Homogeneity of trigonometric equations. — Xo geometric 
meaning can be attached to the result obtained by adding 
an area and a length : or by adding a volume and a length ; 
or by adding a volume and an area. And any equation 
involving terms of different dimensions is also without 
meaning. Hence in every trigonometric equation, each 
term must have the same dimensions. Equations in which 
this condition is fulnlled are called homogeneous equations. 
Therefore every trigonometric equation must be a homo- 
geneous equation. 

This property of trigonometric equations is frequently 
of value as a check on the correctness of reduction pro- 
cess -e-s. For example, suppose an investigation of the rela- 
tions of the parts of any figure is undertaken. The first 
equation used must be homogeneous : any other equation 
derived from it should also be homogeneous : if it fails in 
homogeneity a mistake has been made in the process of 
reduction. 

Exercises. — 1. What are the dimensions of each of the 
following expressions : 

a 7 oV jC Sab':" a/V I * 

> 5 de * 8 d a™ 
2. Write an equation containing six terms of different 
form, each term representing a segment : also an equation 
containing six terms, each representing an area. 



APPROXIMATE VALUES OF NUMBERS. 7 

5. Significant figures. — Zeros are not significant figures, 
if they are supplied to a number merely for the purpose of 
putting the decimal point in the right position. Thus in the 
number .00908, the two zeros immediately following the 
decimal point are not significant figures, but the remaining 
figures, 9, 0, and 8, are significant figures. 

Also, in writing approximate values of large numbers, 
zeros are frequently supplied when necessary to get the 
first significant figure of the number in the proper position 
with reference to the unit's place. Thus in the number 
92,500,000 miles, the 5 is understood to be the nearest fig- 
ure in the hundred-thousands place ; the zeros following 
the 5 indicate a lack of knowledge as to the value of the 
figures in the ten-thousands place and in lower orders ; the 
9, 2, and 5 are the only significant figures in this number. 

6. Errors in approximate values of numbers. — Suppose an 
assumed value of a number, expressed with four significant 
figures, the fourth figure being the nearest figure in that 
place. The ratio of the error to the assumed value of the 
number is not greater than 1 to 2000. For the most un- 
favorable case is when 1000 is assumed as the value of 
1000.5, in which case the ratio is 1 to 2000. 

In a similar manner the ratio of the error of an approxi- 
mate value of a number of five significant figures (the fifth 
figure being the nearest figure) cannot be greater than 1 to 
20,000; and soon. 

7. Errors arising in the processes of calculation. — Suppose 
that assumed values of numbers are introduced into a 
numerical calculation. From the principles of arithmetic 
all numbers arising subsequently in the process of calcula- 
tion will in general have about the same degree of precision 
as was contained in the assumed value of the least degree of 
precision which entered into the calculation. Thus, con- 
sider the following examples : 



8 INTRODUCTION. 

Multiplication. — Suppose it is desired to multiply the 
approximate value .7862 by the number 43.59. The num- 
ber of which .7862 is the approximate value is between 
.78615 and .78625. Multiply each separately by 43.59. 
The products are 34.2682785 and 34.2726375. It will be 
seen that 34.27 will represent either product to four signifi- 
cant figures. Xow multiply .7862 by 43.59 by 

•7862 £ rs t multiplying by 40. then by 3, then by .5. 

43.59 then by .09. 

31.448J0 To reject all figures to the right of the 

2.358 6 vertical line drawn through the products, use 



.393 
.070 



10 all the figures of the multiplicand when multi- 
758 plying by 40, and put in the decimal point ; in 



of o-q _^~g multiplying by 3, strike out the 2 of the multi- 
plicand, but use it to obtain figures which may 
affect the column to the left of the vertical 
line; thus, 3 times 2 are 6. and 6 being more 
than 5, 1 is carried : 3 times 6 are 18. and 1 to 
carry is 19, and so on. In multiplying by 5. 
the 6 of the multiplicand is struck out, but is 
used to obtain 3 to carry ; thus, 5 times 6 are 
30, 5 times 8 are 40, and 3 to carry are 43. and 



31.448 

2 359 

393 

71 



34.27 so on. In multiplying by 9, the 8 is struck 

out, but the 86 now struck out being nearly 90, 
we proceed : 9 times 6 are 54, 9 times 8 are 72, and 5 to 
carry are 77, 9 times 7 are 63, and 8 to carry are 71 ; or 
simply considering the 86 thrown away to be 90. we say 9 
times 9 are 81, 9 times 7 are 63, and 8 to carry are 71. In 
adding the partial products, the right-hand column adds to 
21 ; reject the 1 and carry the 2 to the next column, as only 
four significant figures of the product can be correctly 
obtained. The method is expressed in the following 

Rule. — Use the figures of the multiplier in reverse order. 
Supply the decimal point in the first partial product. Reject 
all unnecessary figures. 



CONTRACTIONS IN COMPUTATIONS. 



9 



Division. — To reverse the process, divide the product 
34.27 by 78.62. 



(43.59 



COMMON 

62)34.27 
3144 


METHOD. 

00000(43.589 
8 


CONTRACTED 

.7862)34.27 
3145 


2 82 
2 35 


20 
86 


2 82 
2 36 

46 
39 

7 


46 
39 


340 
310 


7 
6 


0300 

2896 




74040 

70758 



Crosses are added to make the number of decimal places 
in the dividend equal to the number in the divisor. After 
the multiplication by the quotient figure 4, put as many 
crosses after the 4 as there are unused crosses in the divi- 
dend. Then supply the decimal point. The rest of the 
process is similar to the process of multiplication. 

The square root of .7862. 



COMMON METHOD. 

.78620000(.88667 
64 



CONTRACTED METHOD. 



168 : 


L462 
L344 


1766 


118 
105 


00 
96 


17726 


12 
10 


0400 
6356 


17732 


1 


4044 



.7862(.8867 
64 



168 



176 



1462 
1344 



118 
106 

12 



10 INTRODUCTION. 

A consideration of these examples will show that if data 
are correct to only three significant figures (say an error of 
1 in 1000), the calculation may be kept in four significant 
figures ; for the result will be as precise as the data, and 
further figures are unnecessary and unreliable. 

Similarly, if the data are correct to four significant figures 
(say an error of 1 in 10,000), the calculation may be kept in 
five significant figures. And so on. 

Applications. — The summation 
of series is one of the many prac- 
tical applications of approximate 
values of numbers. Thus, sup- 
pose it is desired to compute to 
ten decimal places the value of 
the series : 

e = 1+ i + i i 2 + rriT3 +etc -' 

in which each term is obtained 
from the one preceding by divi- 
sion, the divisors increasing by 1 
each time. The work proceeds as 
in the margin. It will be seen 
that the nearest tenth decimal 
figure is 5. 

Problems. — Perform the following multiplications, keep- 
ing only the figures which affect the fourth significant figure 
of the product. After each multiplication, perform the re- 
verse operation of division or square root : 

231.1 x 4.414 = 1040 1.414 x 1.414 = 2.000 

.4567 x 9.009 = 4.114 .7454 x .7454 = .5000 

7060 x .0302 = 213.2 57.33 x 57.33 = 3287 

.7854 x 34.25 = 26.90 1.012 x 1.012 = 1.024 



1)1 




2)1 




3) 


.5 


4) 


.166,666,666,67 


5) 


41,666,666,67 


6) 


8,333,333,33 


7) 


1,388,888,89 


8) 


198,412,70 


9) 


24,801,59 


10) 


2,755,73 


11) 


275,57 


12) 


25,05 


13) 


2,09 


14) 


16 




1 


2.718,281,828,46 



CHAPTER I. 



FUNCTIONS OF ANGLES. 




Fig. 11. 



8. Generation of an angle. — The angle BAG may be con- 
ceived to be generated by a turning of a half -line from 
either side of the angle 
to the other side. Thus, 
the position AC may be 
reached by a turning of a 
half -line from AB through 
an angle x + any num- 
ber of revolutions ; or 
the same position may 
be reached by a turning 

from AB in an opposite direction through an angle 
360° — x + any number of revolutions. Also, if b, the 
length of AC, is regarded as negative, the turning from 
AB is either 180° -\-x or a downward turning of 180°— x. 
For example, if, in Fig. 11, x is 30° and b is 10 inches, the 
point G is determined by any of the following values of b 
and x : 

b = + 10; x ==30°, 390°, 750°, 1110°, etc. 

b = - 10 ; x = 210°, 570°, 930°, 1290°, etc. 

b = + 10; a =-330°, -690°, -1050°, etc. 

b = -10; a = -150°, -510°, -870°, etc. 

The half -line AB from which angles will be conceived to 
be measured will be called the initial line ; the final position 
AC of the revolving half-line will be called the terminal 

11 



12 



FUNCTIONS OF ANGLES. 



line of the angle ; the length, b, from the vertex of the angle 
to any point C on the terminal line will be called the 
distance of C. 

The direction of the initial line is entirely arbitrary. It 
will be drawn to the right in the discussions in this chapter. 
The angle will be regarded as positive when generated by a 
half-line turning from the initial line upward to the left, 
as indicated by the arrow, Fig. 11. It will be regarded as 
negative when generated by a half-line turning in the oppo- 
site direction. The distance will be regarded as positive 
when laid off from the vertex of the angle along the 
terminal line. 



9. The four quadrants. — When a positive angle is under 
consideration, it will be conceived to be moved so that 
its vertex is brought at A, 
one side coincides with AB, 
and the angle extends in a 
positive direction. Then if 
the terminal line falls in 
the angle BAD (where DD 1 
is perpendicular to BAB^), 
the angle is said to be of 
the first quadrant; if the 
terminal line falls in DAB 1} 
the angle is of the second 
quadrant; if the terminal 
line is in B X AD 1} the angle 
is of the third quadrant ; if the terminal line is in D X AB, 
the angle is of the fourth quadrant. Thus, 

30°, 390°, -330°, -690° are angles of the first quadrant. 

150°, 510°, -210°, —570° are angles of the second quadrant. 




210°, 570°, -150 c 
330°, 690°, - 30 c 



•510° are angles of the third quadrant. 
390° are angles of the fourth quadrant. 



DEFINITION OF FUNCTION. 



13 



c 

Fig. 13. 



Abscissa and ordinate. — From any point C drop a perpen- 
dicular upon the initial 
line AB. The distance 
from A to the foot of 
this perpendicular is 
called the abscissa of 
C. The length of the 
perpendicular is called 
the ordinate of C. 

Abscissae measured to 
the right of A will be 
considered positive, and 
measured to the left, 
negative. Ordinates 

measured upward from 
the initial line (or ini- 
tial line drawn back- 
ward) will be considered 
positive, and measured 
downward, negative. 
Thus, 



B, 



c 3 



c. 



In Quadrant I, a 
In Quadrant II, a 
In Quadrant III, a 
In Quadrant IV, a 



Fig. 14. 

is + and c is + 
is + and c is — , 
is — and c is — . 
is — and c is -f . 



On the half-line AB, a is zero and c is + . 
On the half-line AD, a is -f and c is zero. 
On the half-line AB Xi a is zero and c is — . 
On the half-line AD lf a is — and c is zero. 



Definition of function. — Suppose a magnitude m, and a 
number of other magnitudes, p, q, r, so related to m, that 
when m changes in value, all of the others change in value ; 



14 



FUNCTIONS OF ANGLES. 



and when m remains fixed in value, all of the others remain 
fixed. In such a case, p, q, r are said to be functions of m. 

10. The six ratios. — If we take in pairs the ordinate, 
distance, and abscissa of a point on the terminal line of an 
angle A, six different ratios can be obtained : 



b b 



It will now be shown that these six ratios are functions of 
the angle A. 

For suppose, first, that b remains the same length while 
the angle A changes in 
magnitude ; say A in- 
creases. Then if A is an 
angle of the first quadrant, 
the point C will move 
upward to the left, its or- 
dinate will increase, and 
its abscissa will decrease. 

Hence -, -, and - will in- 
b c c 

crease, and their reciprocal ratios will decrease. 

In a similar manner by assuming a or c to remain the 
same length, or by assuming the moving point C to be in 
any other quadrant, the same thing may be shown. 

Assume now that C moves so that the angle A is un- 
changed. If C moves either toward A or away from it to 
a point C, the similar tri- 




angles 


ABC and 


AB' 


give : 






BC: 


AC = B'C 


AC 


BC: 


AB = B'C 


AB' 


AC: 


AB = AC 


AB' 



the reciprocal ratios also 
being equal in pairs. 




NAMES OF THE RATIOS. 



15 



Names of the ratios. — For any point on the terminal line 
of an angle, 

The ratio of the ordinate to the distance is called the 
sine of the angle. The sine of x is written sinx. 

The ratio of the ordinate to the abscissa is called the 
tangent of the angle. The tangent of x is written tan x. 

The ratio of the distance to the abscissa is called the 
secant of the angle. The secant of x is written sec x. 

These three ratios connect the ordinate, distance, and 
abscissa of a point ; but it is often convenient to use the 
reciprocal ratios. 

Imagine the triangle ABC X to be taken up from the plane 
and placed so that the vertex C x falls at A, and the side 
CiB coincides with the initial line. B takes the position 




B\ and A the position A'. Then the ordinate of A' will 
equal the abscissa of d, and the abscissa of A' will equal 
the ordinate of C v The three reciprocal ratios are : 



16 FUNCTIONS OP ANGLES. 

First. The ratio of the distance of C x (distance of A') tc 
the ordinate of C r (abscissa of A 1 ) is the secant of B'AA' 
(=€,= 1/). 

Second. The ratio of the abscissa of C\ (ordinate of A') 
to the ordinate of C x (abscissa of A) is the tangent of y. 

Th ird. The ratio of the abscissa of C, (ordinate of A 1 ) to 
the distance of C\ (distance of A') is the sine of y. 

Xow the angle C x (= BAA' = y) is the complement of 
the angle BAC\ ( = x < : that is. y = 90°— x: on this account 
any function of y is called the function of the complement 
of x or the complement's function, or simply cofunction of as* 

Although the prefix co was obtained by thinking of y, 
and therefore x as being acute, the functions of y or the 
cofunctions of x will be defined as applying to all angles. 

The co-named functions. — Tor any point on the terminal 
line of an angle : 

The ratio of the distance to the ordinate is called the 
cosecant of the angle. The cosecant of x is written esc x. 

The ratio of the abscissa to the ordinate is called the 
cotangent of the angle. The cotangent of x is written cot x. 

The ratio of the abscissa to the distance is called the 
cosine of the angle. The cosine of x is written cos x. 

TVhen the group of functions, esc x. cot x. cos x. are 
referred to. they will be spoken of as the co-named func- 
tions of x. 

But it must be noticed that sin;/:, tana;, sec;/; are cofunc- 
tions ; that is. sin x is the cofunction of cos x. just as cos x 
is the cofunction of sin*. In the same way tanas and 
cot x are cofunctions of each other ; sec x and esc x are 
also cofunctions of each other. 

* Xote that the complement of the complement of an angle is the 
angle itself [90~ —(00° — ;<•)=£], and that any co-cof unction of the 
angle would be the function itself. 



RELATIONS BETWEEN FUNCTIONS. 



17 



RELATIONS BETWEEN FUNCTIONS AND SEGMENTS. 
11. In each quadrant 

ordinate of C . distance of C 



distance of C ordinate of C 

That is, sin x esc x=l) 

Similarly, tan x cot x = 1 

sec x cos x = 1 



1. 



(1) 



Therefore sin x is the reciprocal of esc x, esc x is the 
reciprocal of sin x ; for this reason, sin x and esc x will be 
called reciprocal functions of x; similar statements may be 
made of the other functions in group (1). 

In each quadrant, 

ordinate of C _ ordinate of C abscissa of C 
distance of C abscissa of C distance of C 



That is, 
Similarly, 



sin x = tan x cos x 
tan x = sec x sin x 
sec x = esc x tan x 
cso # = cot x sec x 
cot x = cos x esc x 
cos x = sin x cot x 



(2) 



Solving the equations in group (2) for each function, 
tan x cos x cot x sec x ) 



smx = 



sec x cot x 



CSC£ 



sec x sm # 
tan a; = = ; cot x = 



cos x tan x 
cos # esc x 



sec a; 



esc x cos x 
esc a; tan a; 



cot x sin a; 



cos a; 



sm x sec x 
sin x cot a; 



(3) 



tan x esc # J 



18 



FUNCTIONS OF ANGLES. 




Value of each function. 

If x is an angle of the 
first quadrant, the defini- 
tions give 

a b ) 

sin x = - : esc x = - 
b a 

tan x = -\ cot x = - \ 
c a 

b c 

sec x = - : cos x = T 
c b 



(4) 



For the second quadrant, 



sin x 



a b b 

+ =-; csca= - = + - 

6 ' a a 



a a f ,_"~ c c 

— c c ' ' a a 



sec # = = 

— c 



6 -c 

- ; cos x = —=— 
c b 



(5) 



And similarly for the other two quadrants. 

Values of the ordinate, distance, and abscissa of any point 
on the terminal line of an angle. — Clearing the equations in 
(4) of fractions gives for the first quadrant, 

a = b sin x = c tan x ' 

b = c sec x = a esc x ■ (6) 

c = a cot x = b cos x 

The group in (5) gives for the second quadrant, 

a = b sin x = — c tan x 
b = — c sec x = a esc x ■ (7) 

c = — a cot x* = — b cos # 
And similarly for the other two quadrants. 



MNEMONIC DEVICE. 



19 



sec* 



tan* 



sin* 



12. Device for memorizing relations. — The diagram, Fig. 
19, with, the rules following, will be found useful in remem- 
bering the relations 
given in the groups of 
equations (1) to (7). In 
the figure, the angle x is 
drawn for the first quad- esc* 
rant; but if the proper 
algebraic signs of a and 
c are supplied, the dia- 
gram and all the rules 
apply to an angle of 
any quadrant. 

It will be seen that three cross-lines are drawn through 
a and b, a and c, b and c. Starting with the right angle 
and going in the positive direction of angles, the cross-lines 
are marked sin a?, tana;, sec a;, esc a;, cotcc, cos a;, respectively. 

By going along the cross-lines, the relations in groups 
(4), (5), (6), and (7) are given by the two following rules: 




cot* 



cos* 



Fig. 19. 



Rule I. Any segment is equal to the product of the two adja- 
cents ; thus along the horizontal cross-line, a = b sin x and 
b == a esc x. 



Rule II. Any segment or function is equal to the quotient of 
the adjacent by the next ; thus along the horizontal cross-line, 



a b i a b 

sin x = -, a = , o = , esc x = -• 

b esc x sin x a 



By going around either way outside the triangle, consider- 
ing only the functions, equations (2) and (3) are given by 
two rules : 



Rule III. Any function is equal to the product of the two 
adjacents ; thus sin x = tan x cos x. 



20 FUNCTIONS OF ANGLES. 

Rule IV. Any function is equal to the quotient of the adja- 
cent by the next ; thus 

tan x cos x 

sin x = or sin x = 

sec x cot x 

Considering functions on the same cross-line, group (1) is 
given by 

Eule V. Any function is equal to the reciprocal of its oppo- 
site function ; thus sin x esc x = 1. 

ABSOLUTE VALUES AND 
ALGEBRAIC SIGNS OF FUNCTIONS. 

13. Definitions. The numerical value of a magnitude is 
the ratio of the magnitude to another magnitude of the 
same kind taken as a unit. The absolute value of a magni- 
tude is the numerical value of the magnitude considered 
without regard to sign. Thus the numerical value of 
— 5 feet is — 5, and the absolute value is simply 5. 

Absolute values of functions of angles in terms of functions of 
angles between 0° and 90°. — In Fig. 17, denote the positive 
acute value of angle BACi by x. Let the least positive 
value of angle BAC 2 = 180° - x, of angle BAC 3 = 180° + x, 
and of angle BAC 4 = 360° — x. Also make the distances 
AC X , AC 2 , AC 3 , AC 4 all equal. Then the triangles BAC^ 
B 1 AC 2 , B X AC 3 , BAQ are all equal. 

When the revolving half-line is at B or B ±i the angles 
generated from AB are all included in the expression n times 
90°, where n is an even whole number. And when the 
revolving half-line is at D or D ly the angles generated from 
AB are all included in the expression n times 90°, where n 
is an odd number. Hence the angles generated when the 
terminal line is at AC lf AC 2 , AC 3 , or AC A are included in the 
expressions 90 n ± x where n is even, and 90 n ± y where n 
is odd. 



REDUCTION TO FIRST QUADRANT. 



21 



From the equality of the various triangles the ordinates 
and abscissae of Ci, C 2 , C 3 , and C 4 are all equal in absolute 
value. Hence the functions of BAG L , BAC 2 , BAC 3 , BAC 4 




are equal in absolute value to the same functions of x; and 
therefore also equal to the cof unctions of y. That is : 

I. The absolute value of any function of an angle 90 n ± x 
where n is even, is the same as the absolute value of that func- 
tion of x. 

II. The absolute value of any function of an angle 90 n ± / 
where n is odd is the same as the absolute value of that co- 
function Of y. 

Expressed in a formula, / standing for any function : 

f(90n ± x)= ± fx when n is even = ± cofx when n is odd. 

For example : Any function of 150°, 210°, 330°, 390°, 510°, 
— 30°, — 150°, etc., is equal in absolute value to the same 
function of 30° or the co-named function of 60°. 



22 FUNCTIONS OF ANGLES. 

The algebraic signs of the functions may be remembered by 

this device : 

Write the functions in the order 

sin tan sec esc cot cos 
2 3 4 2 3 4 
The functions with positive values are : 

In Quadrant I. All of them. 
In Quadrant II. sin esc 

In Quadrant III. tan cot 
In Quadrant IV. sec cos 
In all other cases the functions are negative in value. 

The method of obtaining the value of a function of any angle 

in terms of an angle between 0° and 90° is given in these 
two rules : 

Eule I. Divide the angle by 90°. If the quotient is even, 
take the same function of the remainder. If the quotient is odd, 
take the cofunction of the remainder. 

Eule II. Prefix the proper algebraic sign. 

Thus sin 172°= cos 82°. esc 274° 9' = - sec 4° 9' 
tan 305° = - cot 35°. cot 329° 51' = - tan 59° 51' 
sec 595° = - sec 55°. cos 935° 22' = - cos 35° 22' 

Notice that the quadrant is indicated by 1 + the quotient. 

14. Functions of 180° ± x, of 90° + x, and of - jr. — Con- 
sider any function, say the sine, of the two angles 180° — x 
and x. It has been shown that sin (180° — x) and sin x are 
equal in absolute value. Consider now their algebraic signs. 

If x is of the first quadrant, 180° — a; is of the second 
quadrant ; then sin x and sin (180° — x) are both + . 



REDUCTION TO FIRST QUADRANT. 



23 



If x is of the second quadrant, 180° — x is of the first 
quadrant ; then sin a and sin (180° — x) are both +. 

If x is of the third 
quadrant, 180° — x is of 
the fourth quadrant ; 
then sin a; and sin 
(180° -x) are both -. 

If x is of the fourth 
quadrant, 180° — x is 
of the third quadrant; 
then sin a and sin 
(180° -x) are both -. 

Hence the equation 
sin (180° — x) = sin x is 
true for all values of x. 

In a similar manner, it may be shown that the equation 
sin (90° + x) = cos x is true for all values of x. Also that 
sin (180° -f- x) = sin (— x) = — sin# are true for all values 
of x. 

In a similar manner, other functions of these angles may 
be considered. The results are shown in the following table : 




sin (180° — x) = sin x. 
tan (180° -*) = - tana, 
sec (180° — x) — — sec x. 
esc (180° — x) = esc x. 
cot (180° -x)=- cot x. 
cos (180° — x) = — cos x. 



sin (90° + x) = cos x. 
tan (90° + x) = — cot x. 
sec (90° -f- a;) = — esc x. 
esc (90° + x) = sec x. 
cot (90° + x) = — tan x. 
cos (90° + x) = — sin x. 



sin (180° + x) = — sin x. 


sin(- 


-*)=- 


- sin a. 


tan (180° + x) = tana. 


tan (- 


-*)=. 


— tan #. 


sec (180° + x) — — sec x. 


sec(- 


-x) = 


sec x. 


esc (180° -f a) = — esc a. 


csc (- 


.»)=- 


- esc a. 


cot (180° + x) = cot a. 


cot (- 


.*)=- 


- cot x. 


cos (180° + a?) = — cos a. 


cos(- 


-*)= 


cos a. 



24 



FUNCTIONS OF ANGLES. 



THE CURVE OE EACH FUNCTION: 

15. If a point moves so that its distance remains the 
same, the ordinate of the moving point will be proportional 
to the sine of the varying angle ; and the abscissa of the 
moving point win be proportional to the cosine of the angle ; 
for the sine of the angle equals the ordinate divided by the 
distance ; and the cosine of the angle equals the abscissa 
divided by the distance ; and therefore if the distance re- 
mains the same, the sine varies as the ordinate ; and the 
cosine varies as the abscissa. 

Suppose, now, a second point to move in the plane so 
that its abscissa is proportional to the angle, and its ordi- 
nate is proportional to the sine of that angle. As the angle 
changes from 0° to 360°, this second point will trace out the 




Fig. 20. — Curve of sines. 



Curve of cosines. 



curve shown in Fig. 20, called a simple harmonic curve, or 
curve of sines. Or, if the ordinate of the second point was 
proportional to the cosine of the angle, the curve is called 
the curve of cosines. 

If the first point moves in a line at right angles to the 
initial line, the ordinate of this point will be proportional 
to the tangent of the angle, and the distance of this moving 
point will be proportional to the secant of the angle ; now 
if the abscissa of the second moving point was proportional 
to the angle and the ordinate proportional to the tangent 
of the angle, the curve traced out is called the curve of 
tangents (Fig. 21) ; or, if the ordinate of the second point 



CURVES OF FUNCTIONS. 



25 




\ / 

\ / 

\ / 
\ / 
\ / 

\ / 

A 

/ \ 

/ \ 

/ \ 
/ \ 


1 80* 


\ 1 

\ 
\ 

' s / 

\ / 

\ / 

\ / 

A 
/ \ 

/ \ 
/ \ 


360' 


o° 90* 


\ / 

\ / 

A 

/ \ 
/ \ 

/ \ 


270° 


\ / 

\ / 

\ / 

\ / 

\ / 

A 

/ \ 

/ \ 

/ \ 

/ \ 

/ \ 

/ \ 



Fig. 21. — Curve of tangents. Curve of cotangents. 




\ 1 


/ 




\ 


\ 


/ 




\ 


\ 

\ / 
\ / 


/ 
/ 




\ 


\- / 


/ 




\ 


X x/ 


/ 

/ 




\ 


-^ s ^- 


u— - 




^ 


0° 


90° 


180* 


270' 360* 
















X 








\ 






/ \ 


\ 
\ 






/ \ 
/ \ 


\ 






/ \ 


\ 






/ \ 
/ \ 
/ \ 


\ 

\ 

\ 



Fig. 22. — Curve of secants. 



Curve of cosecants. 



26 FUNCTIONS OF ANGLES. 

was proportional to the secant of the angle, the curve is 
called the curve of secants (Fig. 22). 

In a similar manner, the first point may be assumed to 
move parallel to the initial line, and the second point may- 
be made to trace the curve of cosecants (Fig. 22), or the curve 
of cotangents (Fig. 21). 

LIMITING VALUES OF FUNCTIONS. 

16. Since from geometry the hypotenuse of a right tri- 
angle can never be less than either side, the distance of a 
point can never be less than either the abscissa or ordinate 
of the point. Therefore the absolute value of the sine or 
of the cosine of any angle is never more than 1, and the 
absolute value of the secant or of the cosecant of any angle 
is never less than 1. The absolute value of the tangent 
or of the cotangent of angles is without limit. 

THE ANGLES EACH LESS THAN 360° CORRESPONDING 
TO A GIVEN VALUE OF A FUNCTION* 

17. From what has been said, it follows that there is 
but one value of any function of a given angle ; but for a 
given numerical value of any function of the angle, there 
will be 0, 1, or 2 values of the angle between 0° and 360°. 

1. If sin x or cos x is more than 1 or less than — 1, or if 

sec x or esc x is between 1 and — 1, x will be impossible. 

2. If sin x = -f- 1 or esc x = -' r 1, x = 90°. 

If sin # = — 1 or esc x = — 1, x = 270°. 

If sec x = +1 or cos x = -j- 1, x = 0°. 

If sec x = — 1 or cos x = — 1, x = 180°. 

3. In all other cases, two values of x between 0° and 360° 
will be obtained. 

* This section can be more easily followed if the diagrams, Figs. 20, 
21, and 22, are referred to. 



TO FIND AN ANGLE FROM A FUNCTION. 27 

Suppose that sin x = JSf (or esc x = j\ t ), and that x x is 
the corresponding acute value of x. Then x also equals 
180° - x 1 . 

If the given equation is sin x = — JV (or esc x = — JSf) 
and sin x x = JSf (or esc x 1 = JV) ; where a^ is acute, then 

a; = 180° + x x or x = 360° - a^ 

If tan x= N (or cot jr = JV), and a^ is the acute value of x, 
another value of x is 180° -+- x v 

If the given equation is tan x = — JSf (or cot x — — JSf) 
and tan x 1 = JSf (or cot x 1 = JSf) where x 1 is acute, then 

a? = 180°-a; 1 or a; = 360° - x. 

If sec x = N (or cos x = JSf) and a^ is the acute value of x, 
another value of x is 360° — x v 

If the given equation is sec x — — JSf (or cos x = — JV) 
and sec x x = JSf (or cos x x = JSf) where x 1 is acute, then 

x = 180° - x x or x = 180° + x v 

Angle less than 180°. — If the problem of obtaining the 
values of an angle from a given function of the angle is 
still further limited by the condition that the angle shall 
be less than 180° (say an angle of a triangle), then : 

(1) If the given function is a sine or a cosecant with a 
negative value, no value of the angle could be obtained ; 
because all angles between 0° and 180° have a positive sine 
and a positive cosecant. 

(2) If the given function is a sine or a cosecant with a 
positive value, and the given value of the function was pos- 
sible, there will be two supplementary values of the angle ; 
because if x y is the acute value, 180° — x 1 will be the obtuse 
value. If a,*! happens to be 90°, it may be said that the 
two values are equal, or there is but one value. 

(3) If the given function of the required angle is any 
other function than the sine or cosecant, there can be but 




28 FUNCTIONS OF ANGLES. 

one value of the angle less than 180° ; acute if the value of 
the function is positive; obtuse if the value of the func- 
tion is negative. 

If it is known that a required angle was not greater than 
90°, a negative value of any function of the angle would 
indicate that no value of the angle was possible. 



FUNCTIONS OF la; AND 2 x. 

18. Functions of I/. — Bisect 
angle BAC, and draw BE parallel 
to AD. Denote angle BAC by x. 
Then angle E = angle ABE = an- 
gle BAD = i x. Also AE = AB. 

BC 2 = AC 2 - AB 2 . Divide by 
BC 2 . 

1 = esc 2 x — cot 2 x. (8) / 

AB = EA = EC == AC AB ^ 

BD BD BC BC BC 

cot \x = csc x + cot x = 1 + cosa? . (9) 

sm x 

— = — and ^R = ^_. whence — BD = BC 
AD~ AC an EA~ EC" * enCe AE X AD~ AC 

2 cos \ x sin \x= sin x. (10) 

The quotient of (8) and (9) is 

tan i x = csc x — cot x = — -. - (11) 

2 sins v J 

The product of (10) and (11) is 

2sin 2 !£ = l-cosa,\ (12) 

The product of (9) and (10) is 

2cos 2 ia = l + cos£. (13) 

The half difference of (13) and (12) is 

cos 2 \x — sin 2 \ x = cos x. (14) 



Fig. 23. 



FUNCTIONS OF PARTICULAR ANGLES. 29 

The reciprocal of (10) is 

-J sec i- x esc \ x = esc #. (15) 

Functions of 2 a-. — Denote now the angle BAD by x. 
Then from (10), (14), and (15) : 

sin 2 x = 2 sin x cos a;. (16) 

cos 2x = cos 2 as — sin 2 x. (17) 

esc 2 a? = \ sec a; esc x. (18) 

The other functions can then be found in terms of these. 

FUNCTIONS OF PARTICULAR ANGLES. 

19. Functions of 45°, 22° 30', etc. — Let angle ^. = 45°. 
Denote the length of AB by c ; then BC = c 
and AC = VcM^ = 1.4142 c. Hence 

tan 45°= cot 45° = 1. 

sec 45° = esc 45° = 1.4142. 

Also, sin 45° = cos 45° = .7071. A- 

Fig. 24. 
From these values of the functions of 45°, 

the values of the functions of 22° 30' can be found by means 
of formulae (9) to (13). Then from the values of the func- 
tions of 22° 30', the functions of 11° 15' can be found; and 
so on. 

Functions of 60°, 30°, 15°, etc.— The tri- C 

angle having angles of 90°, 60°, and 30° is 
one-half of an equilateral triangle. 

Take BC = a : then AC =2 a and 




AB = V4 a 2 -a 2 = 1.7321 a. ^\ 

Fig. 25. 
Hence the functions of 60° and 30° can 

be computed. And, therefore, the functions of 15°, 7° 30', 

etc., may be computed. 



30 



FUNCTIONS OF ANGLES. 



Functions of 72°, 36°, 18°, etc. — Let AC = CD, and DE 

bisect angle D, CB bisect angle (7; take the value of angle A 
to be 72°. All the other angles on the fig- 
ure can then be found. From the equality 
of angles, AD = ED = EC ; also the tri- 
angle A CD is similar to the triangle ADE. 

Denote the length of AC by b, and the 
length of AB=b cos 72° by bx. Then 
AE = b - 2 bx, 

From the similar triangles A CD and 
ADE, 

AC AD __ b 2bx 

AD 




AE 



or 



2bx b-2bx 



.3090. 



Solving : x = cos 72° = sin 18° = \ ( V 5 - 1) 

Denote the length of BC by by ; then 

BC 2 = AC 2 - AB- or b-y- = b~ - tfx 2 . 

That is, y- = 1 - x 2 or sin 2 72° = cos 2 18° = 1 - .3090 2 . 

Hence the other functions of 72° and 18° may be found. 
Also the functions of 36°, 9°, 4° 30 f ; etc, may be found. 

Exercises. — For all values of x, prove that 

sin 2 x -f- cos 2 a; = 1, 
sec 2 x — tan 2 x = 1, 
esc 2 x — cot 2 x = 1, 

and explain how all the functions of an angle may be found 
when the value of only one function is known. 



CHAPTER II. 

EXPLANATION OF THE TABLES. 
I. NATURAL VALUES OF FUNCTIONS. 

20. The values given for the functions of angles in The 
Table of Natural Values of Functions, page i to page vii, 
are approximate values to four significant figures at least. 
The last figure given in each value is the nearest figure in 
that place. Thus the value of sin 32°, which is not less 
than .52985 and not greater than .52995, is given on page v, 
line 25, as .5299. 

Any value printed without a decimal point will have one 
or more figures missing. The decimal point and the missing 
figures will be found to the left and slightly above the fig- 
ures printed for the value. Thus page v, line 29, sin 32° 40' 
is .5398, the 398 being printed in the proper place and .5 
being found higher and to the left. 

tan 32° 40' = .6412, sec 32° 40' = 1.1879, etc. 

Given an angle to find any function of the angle — Angles 
between 0° and 45° will be found in the column marked x 
on the left of the page, and angles between 45° and 90° in 
the column marked/ on the right. When the angle is found 
on the left, the names of the functions are given at the top 
of the columns, x being added to the name of each function. 
When the angle is found on the right, the names of the 
functions (marked y) are given at the bottom of the page ; 
thus: 

31 



32 EXPLANATION OF THE TABLES. 



Page v, 


line 39 


sin 34° 20' 


= .5640 


Page vi, 


line 9 


tan 53° 40' 


= 1.3597 


Page iv, 


line 2 


sec 21° 10' 


= 1.0723 


Page vii, 


line 16 


esc 45° 30' 


= 1.4020 


Page iv, 


line 26 


cot 64° 50' 


= .4699 


Page iv, 


line 7 


cos 22° 


= .9272 



To find function of angles which are not multiples of 10'. — 

When we desire the function of an angle, whose value is 
between two angles of the table, a process called interpolation 
is used. 

By reference to any part of the table, it will be seen that 
small changes in the angle produce nearly proportional changes 
in the values of any function of the angle. A four-place table 
usually gives the values of the functions of angles for every 
10'. By assuming that the change in any function is pro- 
portional to small changes in the angle, the value of the 
function of an angle to a single minute can in general be 
found to four significant figures without appreciable error. 
Conversely, if the value of the function is correct to four 
significant figures, the value of the angle can in general be 
found correct to a single minute. 

A five-place table giving the values of the functions for 
each minute can be used to interpolate to 10"; a six-place 
table for every 10" can be interpolated to 1"; a seven-place 
table for every second can be interpolated to .1", and 
so on. 

With every increase in the number of significant figures, 
the degree of precision is increased and also the labor of 
using the tables. If the data of a computation contain 
angles measured to a single minute, four-place tables can 
be used in the computation; if the angles are measured 
to one-tenth second, seven-place tables can be used ; and 
so on. 



INTERPOLATION. 33 

Let /stand for any function. Then if h is a small addition 
to angle x, and z is the change in any function of x due to a 
change h (10') in the angle, it is assumed 

f(x + h)-fx = h 
z k 

or f(x + h)=fx + -z, 

from which the value of a function of an angle x + h can be 
obtained by substitution. Thus : 

sin 34° 23' = sin 34° 20' + .3 (sin 34° 30' - sin 34° 20') 

= .5640 + .3 x .0024 = .5647. 
tan 53° 44' == tan 53° 40' + .4 (tan 53° 50' - tan 53° 40') 

= 1.3597 + .4 x .0083 = 1.3630. 
sec 21° 11' = sec 21° 10' + .1 (sec 21° 20' - sec 21° 10') 

= 1.0723 + .1 x .0013 = 1.0724. 
esc 45° 38' = esc 45° 30' + .8 (esc 45° 40' - esc 45° 30') 

= 1.4020 + .8 x (- .0040) = 1.3988. 
cot 64° 55' = cot 64° 50' + .5 (cot 65° - cot 64° 50') 

= .4699 4- -5 (- .0036) = .4681. 
cos 22° 7' = cos 22° + .7 (cos 22° 10' - cos 22°) 

= .9272 + .7 (- .0011) = .9264. 

It should be noted that all of the co-named functions of 
acute angles decrease as the angle increases. Hence when 
the tabulated value of a co-named function of an acute 
angle is corrected for an addition to the angle, the cor- 
rection will be negative. 

If functions of angles greater that 90° are desired, they 
are obtained by applying the rules on page 22 ; thus (Quad- 
rant II), 



34 EXPLANATION OF THE TABLES. 

sin 155° 12' = cos 6o° 12' = .4195. 
tan 113° 47' = - cot 23° 47' = - 2.2691. 

sec 167° 9' = -esc 77° 9' = -1.0257. 
esc 123= 12' = sec 33 = 12' = 1.2020. 
cot 147 D 56 ' = - tan 57° 56' = - 1.5962. 
cos 131° 35'=- sin 14° 35' = - .7019. 

And in a similar manner for angles of Quadrants III 
and IV. 

Given the value of a function of the angle, to find the value of 
that angle. — The table is arranged to give the value of the 
function when the angle is known. The reverse operation 
of obtaining the angle when the function is known is some- 
what more inconvenient. Assume that we know the sine 
of the required angle ; the given value may be sought either 
in the column marked sin x at the top of the page or in the 
column marked siny at the bottom of the page. If the 
number is found in the column marked sin jr. and it is 
known that the angle is acute, the value of the angle will 
be found in the column marked x on the left of the page. 
If the given value of the sine is found in the column marked 
sin/, the value of the acute angle will be found in the col- 
umn marked /. Any other function will be found in a 
similar manner ; thus : 

Page v. line 27. If sin A = .8450, A (acute) = 57° 40'. 
Page v. line 4. If tan .1 = 1.8418, A (acute) = 61° 30'. 
Page v. line 41. If sec A = 1.2158, A (acute) = 34° 40'. 
Page iv. line 33. If esc A = 2.2543, A (acute) = 26° 20'. 
Page v, line 34. If cot J. = .6619, A (acute) = 56° 30'. 
Page vi. line 27. If cos A = .7735. A (acute) = 39° 20'. 
If the values of the functions of 45° are memorized 
(sin 45° = .7, tan 45° = 1, sec 45° = 1.4), 



INTERPOLATION TO FIND ANGLE. 35 

it will be known whether the given value of the functions 
should be sought in the columns on the right of the page or 
on the left. Thus if sin A = .9112, the value of A must be 
more than 45°, because the sine of an acute angle increases 
as the angle increases, and the sine of 45° is .7. Since the 
angle is more than 45°^ the given value is sought in the 
column marked sin y. 

As before, let / stand for any function, h a small addition 
to angle x, and z a change in any function due to a change 
k (10') in the angle. Then, since small changes in the angle 
are nearly proportional to the changes in any function, 

h_ f(x + h)—fx 

k~~ z 

, f(x + h) - fx . 

x + h = x + — — — '- — — k. 
z 

If sin A = .6479, the acute value of A 

= 40° 20' + - 6479 ~ ,6472 10' = 40° 23'. 
.6494 - .6472 

If tan A = 3.7539, the acute value of A 

= 75 o + 3.7539 - 3.7321 1Q , = 5<D 5 , 

3.7760-3.7321 

If sec A = 1.2190, the acute value of A 

= 34° 50' + 1 ' 2190 ~ 1 - 2183 10' = 34° 53'. 
1.2208 - 1.2183 

If esc A = 3.9984, the acute value of A 

= 14° 20' 4- — 4-0394 .. q, _ -..o oqr 

3.9939 - 4.0394 
If cot^i = 2.2062, the acute value of A 

= 24° 20' + 2 ' 2062 ~ 2 ' 2113 10' = 24° 23'. 
2.1943-2.2113 

If cos A = .5654, the acute value of A 

= 55° 30' + - 5654 - 5664 10' = 55° 34'. 
.5640 - 5664 



36 EXPLANATION OF THE TABLES. 

If both, values of the angle which are less than 360° are 
desired, any value, not acute, can be obtained by a reversal 
of the rules on page 22 ; thus, 

Ifsin^L = .6479, A = 40° 23' or 139° 37'. 

If tan ^4 = 3.7539, A = 75° 5' or 255° 5'. 

If sec A = 1.2190, A = 34° 53' or 325° 7'. 

If esc A = 3.9984, A = 14° 29' or 165° 31'. 

If cot A = 2.2062, A = 24° 23' or 204° 23'. 

If cos A = .5654, A = 55° 34' or 304° 26'. 

Also when the given function is negative : 

.4 = 220° 23' or 319° 37'. 
.4 = 104° 55' or 284° 55'. 
^. = 145° 7' or 214° 53'. 
A = 194° 29' or 345° 31'. 
^ = 155° 37' or 335° 37'. 
A = 124° 26' or 235° 34'. 

It should be noted that if the given value of the sine or 
cosine of an angle is greater than 4- 1 or less than — 1, no 
value of the angle can be obtained ; and if the given value 
of the secant or cosecant of an angle is between + 1 and 
— 1, no value of the angle can be obtained. 

Exercises. — Copy one value in any row of the following 
table. Then find each of the remaining values in the row. 

sin 0°43' sin 179° 17' .0125 cos 89°17' cos270°43' 

sin 10°18' sin 169° 42' .1788 cos 79° 42' cos280°18' 

sin 200° 41' sin339°19' -.3532 cos 110° 41' cos249°19' 

sin 210° 6' sin 329° 54' -.5015 cosl20° 6' cos239°54' 

sin 40° 23' sin 139° 37' .6479 cos 49° 37' cos 310° 23' 



If sin A = 


- .6479, 


If tan A = 


- 3.7539, 


If sec A = 


- 1.2190, 


If esc A = 


- 3.9984, 


If cot J. = 


- 2.2062, 


If cos A = 


- .5654, 



LOGARITHMS. 



37 



tan 84° 15' tan 264° 15' 

tan 74° 18' tan 254° 18' 

tan 157° 13' tan 337° 13' 

tan 141° 33' tan 321° 33' 

tan 135° 19' tan 315° 19' 

sec 22° 34' sec 337° 26' 

sec 63° 12' sec 296° 48' 

sec 99° 11' sec 260° 49' 

sec 128° 43' sec 231° 17' 

sec 107° 47' sec 252° 13' 



9.9310 cot 5° 45' cot 185° 45' 

3.5576 cot 15° 42' cot 195° 42' 

-.4200 cot 112° 47' cot 292° 47' 

-.7940 cot 128° 27' cot 308° 27' 

-.9890 cot 134° 41' cot 314° 41' 

1.0829 esc 67° 26' esc 112° 34' 

2.2179 esc 26° 48' esc 153° 12' 

-6.2659 esc 189° 11' esc 350° 49' 

-1.5988 esc 218° 43' esc 321° 17' 

-3.2742 esc 197° 47' esc 342° 13' 



II. LOGARITHMS OF NUMBERS. 

21. Definition. — - Logarithms are a series of numbers sever- 
ally assigned to the ordinary numbers, and so arranged that 

the multiplication and division 
of ordinary numbers is accom- 
plished by means of the addition 
or subtraction of their loga- 
rithms. 

Many systems of logarithms 
are possible ; but in the simplest 
systems the ordinary numbers 
are powers of a number called a 
base, and the logarithms are the 
corresponding exponents. 

Thus, if the base is 2, the num- 
bers are powers of 2, and the cor- 
responding exponents are called 
logarithms to the base 2. To 
multiply 64 by 16, we add the 
logarithms of 64 and 16 (6 and 4), 
producing 10 ; and 10 is the logarithm of 1024, the product. 



Numbers 


Logarithms 


or 


or 


Powers. 


Exponents. 


1 





2 


1 


4 


2 


8 


3 


16 


4 


32 


5 


64 


6 


128 


7 


256 


8 


512 


9 


1024 


10 



38 EXPLANATION OF THE TABLES. 

To divide 512 by 32, subtract their logarithms (9 and 5), 
giving 4 ; and 4 is the logarithm of 16, the desired quotient. 

To raise 1 to the fifth power, multiply the logarithm of 4 
(2) by the index of the power (5) 3 producing 10, the loga- 
rithm of 1024, the required power. 

To obtain the cube root of 512, divide the logarithm of 
512 (9) by the index of the root (3), giving 3, the logarithm 
of 8, the desired root. 

Apply to logarithms the rules of algebra relating to 
exponents. Then we have 

Log of product = sum of logs of factors. 

Log of quotient = log of dividend - log of divisor. 

Log of power of a number = log of number x index of power, 

Log of root of a number = log of number -f- index of root. 

CHARACTERISTIC AND MANTISSA. 

If we think of integers as being arranged with zeros 
expressed decimally (thus, 7 = 7.0000), all logarithms may 
be said to consist of two parts, — an integral part and a 
fractional part. 

The integral part of a logarithm is called its characteristic. 
The fractional part is called its mantissa. Thus, if 2.1347 
is a logarithm, 2 is its characteristic, and .1347 is its 
mantissa. 

The characteristic may be either positive or negative, but 
the logarithm is so arranged that the mantissa is always 
positive. Thus, if the logarithm was — 2.4178, the number 
3 would be added and subtracted thus, 3 — 2.4178 — 3 = 
.5822 — 3. In this logarithm. — 3 is the characteristic, and 
.5822 is the mantissa. 

Exercises. — Find the characteristic and mantissa of the 
following logarithms : 41, — If. -J, — i 5 Q -. — f. 



CHARACTERISTIC AND MANTISSA. 39 

DENARY LOGARITHMS. 

Although, a practical table of logarithms could be con- 
structed by taking any positive number as the base of the 
system, the common system of logarithms has 10 for its 
base. 

The system whose base is 10 has two principal advan- 
tages : 

(1) The mantissa of a logarithm depends only on the 
figures of the number and their order; or, to express it 
another way, the mantissa of the logarithm of a number 
is independent of the position of the decimal point in the 
number. For example, the numbers 4.57 and 4570 have 
the same mantissa in their logarithms. 

(2) The characteristic of a logarithm depends only on the 
position from the units place of the first significant figure 
of the number ; or, to express it another way, the charac- 
teristic of the logarithm of a number is independent of the 
figures in the number and of the order of the figures. For 
example, the numbers 45.7 and 23.8 have the same charac- 
teristic in their logarithms. 

To prove the first of these principles, say the logarithm of 
4.57 is 0.6599, and that the logarithm of A is a. Then 

100.6599= 457< 1Q « =A 

But 10 3 =1000 10 3 =1000. 

Hence 10 3 - 65 " = 4570. 10 a + 3 = 1000 A. 

It will be noticed that the decimal point of a number is 
moved to the right, when the number is multiplied by an 
integral power of 10. Therefore the logarithm of the prod- 
uct will be increased by this integer (the exponent of the 
multiplier). But adding an integer to a logarithm does not 
affect the fractional part of the logarithm. Hence the man- 
tissa of the logarithm is unchanged. 



40 



EXPLANATION OF THE TABLES. 



In a similar manner, it may be shown that a motion of 
the decimal point to the left will leave the mantissa of its 
logarithm unchanged. 

To prove the second of these principles consider the follow- 
ing table of logarithms, the base of the system being 
10: 

From an inspection of the table it will be seen that all 

numbers between 1000 and 
10,000 (numbers with the 
first significant figure 
three places to the left of 
the units place) will have 
logarithms between 3 and 
4, and hence 3 + a proper 
fraction; and in general 
that numbers with the 
first significant figure n 
places to the left of the 
units place will have their 
logarithms between n and 
n -j- 1? and hence n + a 
proper fraction. Also that all numbers between .0001 and 
.001 (numbers with the first significant figure four places to 
the right of the units place) will have their logarithms 
between — 4 and — 3, and hence — 4 -f- a proper fraction ; 
and in general that numbers with the first significant figure 
>i places to the right of the units place will have their loga- 
rithms between —n and — (n — 1), and hence — n + a 
proper fraction. 

For example : The number 4570 is between 10 3 and 10 4 ; 
hence its logarithm is 3 + a proper fraction. The num- 
ber .00457 is between 10" 3 and 10~ 2 ; hence its logarithm 
is — 3 -j- a proper fraction. 

It follows, from these two principles, that in the con- 
struction of a table of logarithms to the base 10 : 



Numbers. 


Logarithms. 


.0001 


-4 


.001 


-3 


.01 


— 2 


.1 


-1 


1 





10 


1 


100 


2 


1000 


3 


10000 


4 



TO OBTAIN A LOGARITHM. 



41 



(1) The mantissse need be given bnt once for numbers 
having the same figures in the same order. 

(2) The characteristics need not be given, but can be sup- 
plied by the following rule : 

Make the characteristic equal to the number of places by which 
the first significant figure of the number is removed from the 
units place : positive if the first significant figure is to the left of 
the decimal point, negative if to the right. 



THE USE OF THE TABLE OF LOGARITHMS. 

22. To find the logarithm of a number. — The mantissse of 

the logarithms of numbers from to 1000 are printed on 
pages viii and ix of the tables at the back of the book. 
If any mantissa is printed without the decimal point, a 
figure of the mantissa has also been omitted ; the decimal 
point and missing figure will be found above and to the left 
of the figures in the table. 

The first two significant figures of the number will be 
found in the column marked JV, and the third significant 
figure at the top of the columns. If a given number has 
but one significant figure, two zeros are added ; if but two 
significant figures, one zero is added. Thus : 



PAGE VIII. 




PAGE 


IX. 


Line. 


Number. 


Logarithm. 


Line. 


Number. 


Logarithm. 


11 


2.00 


0.3010 


11 


.065 


.8129 - 2 


14 


23.0 


1.3617 


14 


.068 


.8325 - 2 


17 


2.61 


0.4166 


17 


.717 


.8555 - 1 


20 


29.4 


1.4683 


20 


.074 


.8692 - 2 


23 


32.7 


1.5145 


23 


.777 


.8904 - 1 


36 


456 


2.6590 


36 


.009 


.9542 - 3 


38 


4.70 


0.6721 


38 


.920 


.9638 - 1 



42 EXPLANATION OF THE TABLES. 

To find the logarithm of a number of four significant 
figures, the assumption is made that small changes in the 
numbers produce proportional changes in the logarithms. This 
assumption is only approximately true ; that it is nearly 
true may be seen from an inspection of the table. The 
error in the assumption will not, in general, affect the fourth 
figure of the mantissa of the logarithm. 

Thus, let m stand for mantissa and log for logarithm. 
Then 

m log 200.2 = m log 200 + .2 (m log 201 - m log 200) ; 

= .3010 + .2 (.3032 -.3010). 

Therefore log of 200.2 = 2.3014. 
m log .02307 = m log 230 + .7 (m log 231 - m log 230) ; 
= .3617 + .7 (.3636 - .3617). 
Therefore log of .02307 = .3630 - 2. 

Given a logarithm, to find the corresponding number. — Let 
log N= 1.4997. Then (page viii, line 22) the significant 
figures of JV are 316. The characteristic indicates that the 
figure 3 is one figure to the left of the units place ; hence 
the figure 1 is in the units place and JSF= 31.6. 

Let log N= .5453 — 2. The significant figures of j^are 
351, and since the 3 is two figures to the right of the units 
place, N= 0.0351. 

Let log N= 2.3014. Then 

N=2Q 3014 -.3010 
.3032 - .3010 

Let log N= .3630 - 2. Then 

^Y= ,023Q - 3630 ~- 3617 = .02307. 
.3636 - .3617 



INTERPOLATION. 43 

By assuming either column in this table find the other. 



Number. 


Logarithm. 


Number. 


Logarithm. 


10.62 
1.384 
1886 
273.1 


1.0261 
0.1411 
3.2755 
2.4363 


4.426 

56.81 
6.282 
733.5 


0.6460 
1.7544 
0.7981 
2.8654 



Other methods of writing negative characteristics. — To avoid 
printing negative characteristics, it is usual to arbitrarily 
increase the characteristic by 10. Thus, log of .01657 = 
.2193 — 2 will sometimes be written 8.2193, which is the 
logarithm of 165,700,000, or just 10,000,000,000 times as 
great as the real value. In such a case, L will be used 
instead of log. Thus 

log .01657 = 8.2193 -10 and L. 01657 = 8.2193. 

Suppose the cube root of .01657 is desired. 

Since log .01657 = .2193 - 2 or 8.2193 - 10, 



log V .01657 = .0731 - .6667 or 2.7398 - 3.3333. 

To avoid this negative mantissa, the logarithm of .01657 
is arranged before division, thus : 

log .01657 = 1.2193 - 3 or 28.2193 - 30. 
Therefore log v.01657 = .4064 - 1 or 9.4064 - 10. 

That is : In taking the root of a proper fraction by loga- 
rithms, see that the negative part of the logarithm is divisi- 
ble by the index of the root. 

Negative numbers. — The sufhx n placed after a logarithm 
indicates that the number corresponding to the logarithm is 
negative. If, in obtaining the product of several numbers, 
there is an odd number of negative factors, the suffix n will 



44 



EXPLANATION OF THE TABLES. 



occur an odd number of times ; therefore, by trie principles 
of algebra, the logarithm of the product should have the suf- 
fix it. But if there is an even number of negative factors, 
the suffix a will appear an even number of times ; therefore 
the logarithm of the product will not contain the suffix n. 

In a similar manner, the rules of algebra will serve as a 
guide in the use of the suffix n, in the operations of division, 
involution, and evolution. 



1. Compute 

k = 46.12x .7827. 

log 46.12= 1.6639 

log .7827= 9.8936-10 



4. Compute 



log x = 11.55 1 5 — 10 
x = 36.1 



2. Compute 

x = 525.1 -=- 78530. 

log 525.1 = 12.72n2 -10 
log 78530= 4.8950 



loga:= 7.8252-10 
x = 0.006687 

3. Compute 

x = 5850 -h 0.04754. 

log 5850 = 13.7672 -10 
log .04754 = 8.6771-10 
log x = 5.0901 
x = 123100 



162.4 x 238.9 x 115.5 



N 516.8 

log 162.4 = 2.2106 
log 238.9 = 2.3782 
log 115.5 = 2.0626 

6.6514 

log 516.8 = 2.7133 

log x 3 = 3.9381 
log x = 1.9690 

" x = 93.12 

>. Compute 



X =\ 



3 129.97 x .0499 



log 129.97= 2.1138 
log .0499= 8.6981-10 



10.8119 - 10 
log 278.7= 2.4451 



log x 3 = 28.3668 - 30 
x = 9.455i 
x = 0.2855 



log x = 9.4556 - 10 



Exercises. — I se problems on page 10. 



EXERCISES. 



45 



III. LOGARITHMS OF FUNCTIONS. 

23. The table, pages x to xvi, Logarithms of Functions, 
contains the logarithms of the natural functions of angles 
from 0° to 90°. To avoid printing negative characteristics, 
the characteristics of logarithms of numbers less than 1 
have been printed 10 too large (indicated by the letter L 
in the first, second, and sixth columns). Hence —10 will 
be understood after such logarithms. 

Just as in the other tables, any number printed without 
the decimal point will have also one or more figures missing, 
which figures will be found above and to the left. 

The method of interpolation is the same as in the use of 
the table of natural functions. 

Exercises. — Copy a value in any row in the following 
table ; then find each of the remaining values in the row : 



sin 18° 5' 


sin 161° 55' 


9.4919 


cos 71° 55' 


cos 288° 5' 


sin 26° 6' 


sin 153° 54' 


9.6434 


cos 63° 54' 


cos 296° 6' 


sin 25° 52' 


sin 154° 8' 


9.6398 


cos 64° 8' 


cos 295° 52' 


sin 196° 14' 


sin 343° 46' 


9.4465 n 


cos 106° 14' 


cos 253° 46' 


sin 251° 31' 


sin 288° 29' 


9.9770n 


cos 161° 31' 


cos 198° 29' 



tan 112° 14' tan 292° 14 

tan 101° 46' tan 281° 46 

tan 103° 17' tan 283° 17 

tan 177° 37' tan 357° 37 

tan 34° 44' tan 214° 44 

sec 22° 13' sec 337° 47 

sec 142° 23' sec 217° 37 

sec 76° 8' sec 283° 52 

sec 96° 26' sec 263° 34 

sec 69° 36' sec 290° 24 



0.3885 n cot 157° 46' cot 337° 46' 

0.6813 n cot 168° 14' cot 348° 14' 

0.6269 n cot 166° 43' cot 346° 43' 

8.6193^ cot 92° 23' cot 272° 23' 

9.8409 cot 55° 16' cot 235° 16' 

0.0335 esc 67° 47' esc 112° 13' 

0.2144ri esc 232° 23' esc 307° 37' 

0.6204 esc 13° 52' esc 166° 8' 

0.9506 ft esc 186° 26' esc 353° 34' 

0.4577 esc 20° 24' esc 159° 36' 



CHAPTER III. 

RIGHT TRIANGLES. 

24. The primary object of trigonometry is the investiga- 
tion of the methods of computing the numerical values of 
the sides and angles of plane triangles. When, from any 
data, such numerical values have been obtained, the triangle 
will be said to be solved. 

With reference to the number of possible answers, a 
problem is said to be indeterminate, ambiguous, determinate, or 
impossible : 

(1) When there is an infinite number of answers, the 
problem is indeterminate. An algebraic illustration is : 
find x and y from the equation x — y = 3. 

(2) When there is more than one answer, but. not an 
infinite number of answers, the problem is ambiguous. Tor 
example : find x from the equation x 2 -f- 6 = 5 x. 

(3) When there is one answer and but one answer for 
each unknown, the problem is determinate. Tor example : 
find x and y from the equations x-\-y=7 and x — y = 3. 

(4) When there is no answer, the problem is impossible. 
Tor example : find x from the equation 

2 x—5 = negative square root of (x 2 — 7). 

In trigonometry, negative and imaginary answers are, in 
general, rejected, as not indicating true solutions. 

It is known from geometry that at least three indepen- 
dent conditions are necessary to make the solution of a 

46 




RELATIONS OF ANGLES AND SIDES. 47 

triangle determinate.* In most cases, three snch conditions 
are sufficient. 

Two classes of triangles will be considered, — right and 
oblique. Any oblique tri- 
angle can be divided into 
two right triangles by let- 
ting fall from any vertex 
a perpendicular upon the 
opposite side. Hence the 
solution of all triangles 
depends upon the solution 
of right triangles. Fig. 27. 

25. In a right triangle, let A denote the value of angle A, 
B denote the right angle, and C the value of angle C. Also 
let a and c denote the lengths of the sides of the triangle 
opposite A and C, respectively, and b denote the length of 
the hypotenuse. 

Relations of angles. — In a right triangle, one of the three 
conditions necessary to determine its size and its shape is 
explicitly given. For the angle B = 90°. Therefore 

A+ C=90°; ^ = 90°-C; C=90°-A. 

Hence, when either acute angle of a right triangle is given, 
the other acute angle is readily found. 

Relations of sides. — The relation existing between the 
sides of a right angle is well known. It is 



b = Vc 2 + a 2 ; c = V (b + a) (b — a) ; a = V(b + c)(b — c). 

Hence, if any two of the sides of a right triangle are 
given, the other side can be computed. 

* This is expressed by saying that the triangle is of triple mani- 
foldness. 



48 EIGHT TRIANGLES. 

Compute a, given b= 145, c = 143. 



a = V(145 + 143) (145 - 143) = 24. 
Compute c, given a = 399, b = 401. 
c 

Exercises. - 



= V(401 + 399)(401-399) = 40. 

Given b and c ; find a in each of the following 



b 


c 


a 


919.2 


918.94 


21.90 


865.1 


860.74 


86.90 


888.9 


888.84 


10.34 


150.0 


149.50 


12.29 


525.1 


523.84 


36.35 



Relations of sides and angles. — Assume AB to be an initial 
line, c the abscissa, a the ordinate, and b the distance of G. 







( 






sec 


A /"^ 


tan A 




b/ 


^/^ 


a 


esc A 






sin A 


/\/ 


c 




B 


cot A. 


cos 


A 





Fig. 28. 



The rules on page 19 give 

Any required side = product of adjacent side and adjacent 
function. 



SOLUTIONS USING NUMBERS. 49 

Any function of A — quotient of the adjacent side by the next 
side. 

If it is desired to use the angle G in a computation, the 
letters a and c, A and C, can be interchanged in Fig. 28, 
and the rules applied to the resulting figure. 



SOLUTIONS. 

26. 1. Given 6 = 120, ^ = 11° 25'; find a. 

The horizontal cross-line in Fig. 28 connects the given side 
b and the required side a; hence by the rule, a = b sin A. 

That is, a = 120 sin 11° 25' - 

= 120 x .1979 = 23.75. 

2. Given c = 50.7, A = 57° 7' ; find b. 

Here we use the vertical cross-line, since that connects 
the given side and the required side. Hence b = c sec A. 

That is, b = 50.7 sec 57° 7' 

= 50.7 x 1.8419 = 93.38. 

3. Given a = 71, A = 65° 32'; find c. 
The oblique cross-line gives 

c = acot^ = 71cot65°32' 
= 71 x .4550 = 32.3. 

4. Given b = 75, a = 43.8 ; find A. 

Here A can be found by means of its sine or its cosecant. 
As the division by 75 is sliglitly easier, we select the sine. 

sin ^ = — =.5840; ^1 = 35° 44'. 

75 



50 



RIGHT TRIANGLES. 



5. Given a = 849, c = 920 ; find A. 

Here the oblique cross-line connects the two given sides. 
As 920 is the easier divisor, 

tan A = ^ = .9228 ; A = 42° 42'. 
920 ' 

6. Given & = 401, c = 399 ; find A. 

Here cos A = ^ = .9950. 

401 

In this solution, A cannot be determined with much pre- 
cision as the table gives 5° 43', 5° 44', or 5° 45'. 
Hence find a first ; a = 40. Then 

esc A = — = 10.025 ; .4 = 5° 43V- 
40 2 

It will be noticed that the cosines and secants of small 
angles change slowly ; and that sines and cosecants of angles 
near 90° change slowly. When precision is desired, such 
functions of these angles are avoided. 

The following table gives the three sides and one angle 
of eight different right triangles. Assume any two parts 
of any one of these triangles and compute another part. 



a 


b 


c 


A 


4 


5 


3 


53° 8' 


12 


13 


5 


67° 23' 


24 


25 


7 


73° 44' 


20 


29 


21 


43° 36' 


12 


37 


35 


18° 55' 


28 


53 


45 


31° 53' 


60 


109 


91 


33° 24' 


84 


205 


187 


24° 11' 



APPLICATIONS. 



51 




Fig. 29. 




Fig. 30. 



PKOBLEMS. 

[Answers to these problems will be found on page 177.] 

27. 1. A tower, BC, stands on flat horizontal ground ; a 
distance of 250 feet (c) was measured from the base of the 
tower to a point A. At this point the angle 
between the horizontal (AB) and the visual 
ray (AC) to the top of the tower was found 
to be A = 22° 20'.* From these data, com- A - 
pate the height of the tower. 

2. A lighthouse, BC, is 87 (a) 
feet high. From the top C, the 
angle between a horizontal line 
CD and the visual ray to a boat 
A is found to be 12° 40'. t From 
these data compute the distance 
of the boat from the base of the 
lighthouse. 

3. Two forces, AB of 240 lbs. and AD of 100 lbs., act 
at right angles upon the same point of a body. Find the 
angle which their resultant AC makes with the larger 
force. 

4. The length of the projection of AC on a line AB is 
23 inches ; the angle which AC makes with this projection 
is 38°. Find the length of AC 

5. A cannon ball, fired in the air at an angle of 33° 3' 
with the horizon, passes over 880 feet in one second. How 
high is it at the end of the second ? $ 

* This angle is called the angle of elevation of the top of the tower, 
as seen from A. 

t This angle is called the angle of depression of the boat as seen 
from C. 

X This increase of height per second is called the vertical component 
of the velocity. 



52 RIGHT TRIANGLES. 

6. In order to measure the breadth of a river, a base line 
AB was measured along the bank, the point B being directly 
opposite a tree C on the other bank. The base line AB was 
96 (c) feet, and the angle BAG was found to be 52° 30' (A). 
What is the breadth of the river ? * 

7. When the rays of the sun make with the horizontal 
plane an angle of 53° 35' (-4),t a vertical pole throws a 
shadow 11.8 (c) feet long. Find the length of the pole if 
its lower end is 1 foot (d) underground. 

8. From a window of a 
house, the angle of ele- 
vation of the top of a ^'~ 
church tower on the op- ^" 
posite side of the street '^*' 
was 56° 36' (A), and the ^ 

angle of depression of A<~ iB 

the base of the tower "^^ 

was 10° 24' (Aj). The ^ ^^ 

street being 60 feet (c) ^-^ 

wide, find the height of „ OH c 

' . & Fig. 31. 

the steeple. 

9. A house is 24 feet (a) high. From the top of the 
house, the angle of elevation of a church tower on the 
opposite side of the street is 36° 54' (A), and the angle 
of depression of the base of the tower is 18° 56' (A x ). Find 
the height of the steeple. 

10. From the top of a hill, the angles of depression of the 
top and of the bottom of a house 44 feet (a) high are found 
to be 28° 4' (.4) and 43° 36' (4), respectively. Find the 
height of the hill above the ground on which the house stands. 

* A segment drawn across an angle is said to subtend the angle. 
Thus in a triangle ABC, BC subtends the angle A. 

t The angle which the sun's rays make with the horizon is equal in 
degrees to the arc which measures the altitude of the sun. 



APPLICATIONS. 53 

11. From the top of a bluff overhanging a river, the 

angle of elevation of the top of a tree on the other side of 

the river is 5° 22' (A), and the angle of depression of the 

bottom of the tree is 3° 40' 

p 

(A x ). The tree is known 

to be 79 feet (a) high. 

What is the breadth of the /' 

river ? y' / 

12. A man sailing due /' / 
north observes two objects ,'' / 
directly west. After sail- ,'' / 
ing 12 miles (a), the direc- 
tions of the objects make ,'' /_ 



B 



angles of 22° 38' (<7)* and D 

of 16° 16' (Ci) with the FlG ' 32 ' 

ship's course. How far apart are the objects ? 

13. At a horizontal distance of 71 feet (c) from the middle 
of the base of a church tower, the angle of elevation of the 
top of the tower = 24° 27' (A), and of the top of the steeple 
= 66° 10' (A,). Find the height of the steeple. 

14. Two observers, 570 (c) feet apart and facing each 
other, observe a balloon in the vertical plane passing 
through them, at angles of elevation of 67° 22' (A) and 
61° 56' (Aj), respectively. What is the height of the bal- 
loon above the ground ? 

15. The angle of elevation of a house on the bank of a 
creek, observed from the opposite bank = 48° 35' (A). The 
observer walks straight back from the river a distance of 
43.15 feet (c), and the angle of elevation is then 33° 24' (A^. 
Find the breadth of the creek. 

* The angle which a horizontal line makes with the meridian is called 
the bearing of the line. Thus if BC is a meridian, and a point A is 
22° 38' west of south from a point C, the hearing of A from C is 
written S. 22° 38' W. 



54 EIGHT TRIANGLES. 

16. The angles of elevation of the top of a steeple at 
distances of 50 feet (c) and 450 feet (c^ are found to be 
complementary. Find the height of the steeple. 

17. A man at the top of a lighthouse observes a boat 
heading straight for the lighthouse ; the angle of depression 
is 34° 40' (A) ; ten minutes (t) later, it is 55° 35 r (AJ. How 
soon will the boat reach the lighthouse ? 

18. Rain falls at an angle of 75° 30' (A) with the ground. 
The wind blows at right angles to a wall, 17.5 feet (a) high. 
How far from the wall can a man stand without getting 
wet, his height being 5.5 feet (h) ? 

19. From the end of the diameter of a circle, a chord is 
drawn, the length of which is n times the diameter. Find 
the angle between the chord and the diameter. Take n— .62. 

20. Find the altitude of the sun if the height of a man is 
n times the length of his shadow. Take n — .74. 

21. From the top of a lighthouse 221 feet (a) above the 

ocean, two boats are ob- 

C 
served, one due west hav- „ ^ 

ing an angle of depression .-' 

of 67° 37' (A), the other / / 

due south having an angle ^' / 

of depression of 74° 49' /' / 

■(D). Find the distance „-"' 

between the boats. ^- 

22. To measure the 
height BC of a tower 
across a river, the tower 
being directly opposite a 
point A, a distance AD ^ 
= 500 feet (d) was meas- 
ured along the bank. At A the angle of elevation of the 
top of the tower was 13° 58' (A), and at D the elevation was 
10° (D). Find the height of the tower. 



/ 
/ 

/ 



Fig. 33. 



AREA. 



55 



THE AREA OF A RIGHT TRIANGLE. 

28. From the right angle of the triangle, drop a perpen- 
dicular upon the hypotenuse. Denote the length of this 

c 




perpendicular by p, and the segments of the hypotenuse by 
d and e. Denote the area by K. 



By geometry, 



K— ^ac = ^bp. 



Exercises. — Prove the following relations involving the 
area of a right triangle : 

(1) 2 K = a 2 cot A. 

(2) 4.K = b 2 sin2A. 



a% 



(3) 2K = c 2 tanA 

(4) K =p 2 csc2A. 



(7) 4K 2 = a 2 (b 2 

(8) 4iT 2 =c 2 (6 2 -c 2 ). 

(9) 4/f =b 2 -(a~c) 2 . 
(10) 16 IO - 4 a 2 d 2 K 2 = a?d 2 . 



(5) 4.K = (c 2 -a 2 )tan2A (11) 2iT=d 2 (tan.4+tan 3 .4). 

(6) 4 K 2 = b 3 d - b 2 d 2 . (12) 8 K= (a + c) 2 - (a - c) 2 . 

(13) 4 iT=(a + b + c) (a - b + c). 

(14) 8 JT= [(a + c) 2 + (a - c) 2 ] sin 2 A 

If iT is eliminated between any two of these relations, a 
relation between the parts of the triangle is obtained. Thus 
if the 2d and 9th relations are used, there is obtained 

& 2 sin2^L = 6 2 -(a-c) 2 . 



56 RIGHT TRIANGLES. 

23. A man having traveled 32 miles in a straight line, 
finds that he has made 17.3 miles more toward the north 
than toward the east. What direction has he been traveling? 

24. A triangular field is at the corner of two roads which 
form a right angle. The area of the field is 32,130 sq. ft., 
and the total length of the fence enclosing it is 918 feet. 
Find the frontage on each road. 

25. A triangular field is at the corner of two roads which 
form a right angle. The area of the field is six-elevenths 
of an acre, and the length of the fence along the two roads 
is 187 feet (the total front fence). Find the length of the 
fence on the back of the field. 

SOLUTIONS USING LOGARITHMS. 



29. 1. Given 


3. Given a = .129, c = .089. 


a = 152 


!, ^L = 18°25'. 


Find A and b. 


Find b and 


c. 


La 9.1106 




log csc ^L 0.5004 


L c 8.9494 




log a 2.1818 
log cot A 0.4776 


A=oo° 24' log tan .4 0.1612 
log sec A 0.2458 


6 = 481.1 


log b 2.6822 


Lc 8.9494 


c=456.5 


log c 2.6594 


b = .1567 Lb 9.1952 


2. Given 
c = .3917, .4 = 65° 14'. 
Find a and b. 

log tan .4 0.3360 
Lc 9.5930 
log sec A 0.3779 


4. Given 

5 = 86.53, A = 56° 3'. 

Find a and c. 

isin^L 9.9188 
log b 1.9372 
Zcos.4 9.7470 


a = .8492 


La 9.9290 


a = 71.78 log a 1.8560 


&=.9352 


L b 9.9709 


c=48.33 logc 1.6842 



SOLUTIONS USING LOGARITHMS. 57 

5. Given a = 50.7, 6 = 93.4. ' Find the unknown parts 
and the area. Check the work. 

6 = 93.4 log 6 1.9703 

a = 50.7 log a 1.7050 

.4 = 32° 53' log esc A 0.2653 

.-. 0=57° 7' log cot A 01894 

log a 1.7050 \ 

c = 78.42 log c 1.8944 

L .5 9.6990 J 

K= 1988 log K 3.2984 

b + a = 144.1 log (6 + a) 2.1587 

6 - a = 42.7 log (6 - a) 1.6304 

Check: logc 2 3.7891 

6. Given 6 = 31.6, c = 31.4 ; find a and A 

Here the hypotenuse and side are given nearly equal. 

By (11), 

tan 1 A = esc A — cot A = — c = \ ~ ° - 

2 a ylb + c 

L(b-c) 9.3010 
log (6 + c) 1.7993 
log a 2 1.1003 

L tan 2 i .4 7.5017 
a = 3.549 log a 0.5501 

i^ = 3°13i r ; ^4 = 6°27' ZtaniJ. 8.7508 

Exercises. — Assume any two parts and find the remain- 
ing parts in the following right triangles : 

1. a = 250 6 = 627 c = 575 A = 23° 30' K= 71880 

2. a = 13.13 6 = 21.94 c = 17.58 A = 36° 45' JT= 115.4 

3. a = 41.54 6 = 42 c = 6.208 ^1 = 81° 30' K= 1289 

4. a = .3864 6 = 4.65 c = 4.634 ^L= 4° 46' K = .8953 



58 RIGHT TRIANGLES. 

PROBLEMS. 

30. Solve the following problems, using logarithms : 

26. A wall is surrounded by a ditch. A ladder placed 
with its foot at the edge of the ditch just reaches the top 
of the wall on the other side of the ditch. The length of 
the ladder was 33.7 feet, and the angle which it formed 
with the horizontal plane was 24°. Find the height of the 
wall and the width of the ditch. 

27. Two forces at right angles, one of which is 7.25 
pounds, are acting at the same point of a body. Their 
resultant divides the angle between them into parts pro- 
portional to 3 and 5, the smaller angle being next to the 
given force. Calculate the other component force and the 
resultant of the two forces. 

28. The diagonal of a sheet of paper is 16.5 inches, and 
it makes, with the shorter side, an angle of 57° 57'. Calcu- 
late the sides of the sheet. 

29. On flat horizontal ground 154 feet from the foot of 
a tower, the angle of elevation of the top was 27° 10'. The 
axis of the telescope was five feet from the ground. Re- 
quired the height of the tower. 

30. The latitude of Philadelphia is 40° X. Taking the 
radius of the earth at Philadelphia to be 3960 miles, find 
the radius and the length of 1° of the parallel passing 
through Philadelphia. 

31. The circumference of the meridian passing through 
Paris is 40,000 kilometers. The length of 1° of the parallel 
passing through Paris = 73.13 km. Find the latitude of 
Paris, considering the earth a sphere. 

32. Two forces, P=7.25 lbs., and Q = 10.3 lbs., act at 
right angles on a body. Calculate the resultant and the 
angle that it makes with P. 



APPLICATIONS. 59 

33. Calculate the velocity per second of a point in 40° 
latitude as the earth rotates on its axis. Take the radius 
of the earth = 3960 miles. 

34. A frigate is 4 miles west from a steamer which is 
sailing north at the rate of 22 miles an hour. How far in 
front of the steamer must the gunner on the frigate aim 
in order to strike the steamer, if the average velocity of the 
cannon ball for that distance is 800 feet per second? 

35. A square is inscribed in another square ; the parts of 
the sides of the outer square made by the corners of the 
inner square are 965 (m) and 603 (n). Find the angle made 
by the diagonal of the inner square and the side of the 
outer square. 

36. An observer sees a cloud in the southwest at an 
angle of elevation of 43° 35'. Another observer 2526 feet (b) 
south of the first sees the same cloud in the northwest. 
How high is the cloud ? 

37. A pedestal 12 feet high (a) supports a column 13 feet 
high Qi). An observer, whose eye is in the same horizontal 
plane with the base of the pedestal, sees the pedestal and 
column each subtending the same angle. What is the hori- 
zontal distance of the observer from the pedestal ? 



CHAPTER IT 



THE ISOSCELES TRIANGLE AND THE REGULAR POLYGON. 



31. The isosceles triangle can be 
divided into two equal right triangles 
by drawing an altitude npon the un- 
equal side. Hence the formulae for 
right triangles will apply to the half- 
triangle AB X C. That is, 

p = b sin A = -L c tan A. 
b = i c sec A = p esc A. 
c = 2 p cot A = 2 b cos A. 



Area 



cp 



^ be sin A 




±b 2 siii2A = \c 2 tanA. 



Fig. 35. 



In these relations, A may be changed to 4 C and the co- 
functions taken. 

Exercises. — Assume any two parts in any row of the fol- 
lowing table and compute the remaining parts in that row : 



p 


b 


c 


A 


K 


16.80 


17.00 


5.200 


81° 12' 


43.68 


1.960 


2.275 


2.310 


59° 29' 


2.264 


70.00 


101.5 


117.0 


43° 36' 


5145 


10.80 


33.30 


63.00 


18° 55' 


340.2 


3.179 


4.261 


5.681 


48° 12' 


9.034 


25.00 


62.70 


115.0 


23° 30' 


1438 


.5878 


1.000 


.1618 


36° 00' 


.4755 


18.08 


83.50 


163.0 


12° 30' 


1474 



60 



APPLICATIONS. 61 



PROBLEMS. 



32. 38. Two trains start at the same station at the same 
time on tracks making an angle of 67° 30'. Each goes with 
a velocity of 27 miles per hour. In how many minutes 
will the distance between them be equal to 5 miles ? 

39. The chord of a circle is 27 inches, and the angle at 
the center subtended by the chord is 116° 12'. Find the 
radius. 

40. The angle at the vertex of an isosceles triangle is 
87° 13', and the sum of the two unequal altitudes is 50. 
Find the base. 

41. The ratio of the altitude from the vertex of an isos- 
celes triangle to one of the equal altitudes is 1.2, and each 
of the equal sides = 5.3. Find the base angle and the 
area. 

42. From a point without a circle 12 feet (d) from the 
center, two tangents are drawn, making with each other an 
angle of 25° 3' (A). Calculate the area of an equilateral 
triangle inscribed in this circle. 

43. From a point without a circle 23.2 feet from the cen- 
ter, two tangents are drawn. The radius of the circle equals 
9.6 feet. Calculate the side of an equilateral triangle whose 
area is equal to that of the figure bounded by the two tan- 
gents and by the smaller arc included between the points of 
contact. 

44. The slant height of a right cone is 13 feet, and it 
makes with the plane of the base an angle of 67° 23'. Find 
the convex surface of the cone. 

45. When a cone was cut through its axis, the area of the 
section was found to be 10 sq. inches and the angle at the 
vertex 40° 30'. Find the convex surface of the cone. 



62 ISOSCELES TRIANGLE AND REGULAR POLYGON. 

46. The convex surface of a cone = 80 sq. inches (S), and 
the angle at the vertex of the principal section = 39° 13' ((7). 

Find the volume of the cone. 

47. When the convex surface of a right cone was spread 

out in a plane, it vras found to be a sector of a circle whose 
central angle vras 108' 36'. and the chord subtending the 
corresponding arc vras found to be 12 inches. Calculate 
the volume of the cone. 

48. The altitude to the base of an isosceles triangle is 
5.4 feet, and one of the equal altitudes is 7.12 feet. Find 
the area. 

49. The area of an isosceles triangle is equal to the area 
of the semicircle on the base as a diameter. Find the angle 
at the base. 

50. In a circle whose radius is 9.4 inches, a chord of 14.4 
inches is drawn. Find the central angle corresponding to 
this chord. 

51. Calculate the area of a circle in which a chord 5 feet 
long subtends a central angle of 77 3 36'. 

52. The lengths of the sides of a rectangle are 7S ft. and 
71 ft. The middle points of its sides are the corners of an 
inscribed rhombus. Find an acute angle of this rhombus. 

53. A point moves in the arc of a semicircle of radius b 
with a uniform velocity of a feet per second. In how many 
seconds is it at a distance of c feet (measured on the chord) 
from its starting point ? Take b = 10. a = 3. c = 16. 

54. Two equal forces act on a body, the angle between 
them being 146° 48'. Their resultant is 12 pounds. Find 
the component force?. 

55. What must be the angle between two equal forces, 
each 3.5 pounds, if their resultant is 1 pound ? 



REGULAR POLYGONS AND STARS. 



63 



THE REGULAR POLYGON. 



33. Denote the number of 
sides by n. Then the half 



angle at the center is 



180 c 



which denote by A. Also de- 
note the side by a, the radius 
of the circumcircle by R, and 
the radius of the incircle by r. 
Then 



r = \ a cot A 



a = 2BsmA = 2r tan A. 
R = r sec A — \a esc A. 
R cos A. 
Area polygon = \ nar 
Area incircle = -n-r 2 = 
Area circumcircle = irR 2 = 




Fig. 36. 



4- no? cot A. 



\ ira 2 cot 2 A. 
i Tret 2 esc 2 A. 



Prove the following properties of regular polygons : 

The area of the ring between the incircle and circumcircle 
of a regular polygon is equal to the area of the circle on 
one side as a, diameter. 

The perimeter of a regular polygon : the perimeter of its 
incircle = area of polygon : area of incircle. 

The area of a regular inscribed polygon of an even num- 
ber of sides is a mean proportional between the areas of the 
regular inscribed polygon and the regular circumscribed 
polygon of half the number of sides. 

Stars. — Divide the circumference of a circle into n equal 
parts. Let s denote a whole number less than %n\ join 
every sth point of division of the circumference (or every 
n — sth point). There is obtained an ?i-pointed star of 
species s. 



64 ISOSCELES TRIANGLE AND REGULAR POLYGON. 



1. If s is 1, the perimeter is continuous and non-inter- 
secting ; the figure consists of a regular polygon of n sides. 

2. If n is a multiple of s, say n = rs, the perimeter is 
discontinuous ; the figure consists of s regular polygons 
each of r sides. For example, the fifteen-pointed star of 
species 3 consists of 3 regular pentagons. 

3. If n and s have a common divisor, say n = ru and 
s = rv, the perimeter is discontinuous ; the figure consists 
of r different stars each of them having u corners. For 
example, the fifteen-pointed star of species 6 consists of 3 
five-pointed stars. 

4. If n and s are prime to each other, the perimeter is 
continuous and intersecting. For example, the fifteen- 
pointed star of species 4. 

Properties. — 1. In a star, prove that the sum of the 
n angles = (n -2 s) 180°. 

2. Prove that a = 2 R sin- 180° = 2 r tan- 180°, 

n n 

B being the radius of the circumcircle, r the radius of the 
incircle, and a the length between any two points joined in 
the construction. 

Exercises. — The following table gives elements of seven 
regular polygons. Assume the number of sides and any 
one element to compute the other elements. 



n 


a 


B 


r 


K 


5 


24.68 


21.00 


16.99 


1049 


6 


20.78 


20.78 


18.00 


1122 


7 


13.88 


16.00 


14.41 


700 


8 


33.14 


43.30 


40.00 


5300 


12 


14.00 


27.00 


26.12 


2195 


20 


6.257 


20.00 


19.75 


1236 


25 


10.90 


43.50 


43.16 


5880 



APPLICATIONS. 65 

PROBLEMS. 

34. 56. The diagonal of a regular pentagon is 4.3 feet (d). 
Find the area of the pentagon. 

57. The area of a circle is 15 sq. feet. Find the area of 
a regular heptagon whose perimeter is equal to the circum- 
ference of the given circle. 

58. Find the largest and smallest diagonals of a regular 
polygon of 7 (2 n + 1) sides, each side being 4 yards (a). 

59. In a park it is desired to make a flower garden to 
contain one acre, and to be of the form of a regular five- 
pointed star. Find each side of the bounding decagon. 

60. Eegular polygons of 27 sides are inscribed in and 
circumscribed about a circle whose radius is 9.2, the sides 
of the two polygons being respectively parallel. What is 
the area of the figure included between the perimeters of 
the two polygons ? 

61. The area of a regular inscribed polygon of n sides 
is K. Find the area of a regular polygon of m sides 
circumscribed about the same circle. Put n = 9, m = 7, 
K= 100. 

62. In a circle whose radius is 2 feet two parallel chords 
are drawn on the same side of the center. One is equal to 
the side of the regular inscribed polygon of 9 sides, the 
other to the side of the regular circumscribed hexagon 
(about the same circle). Calculate the area of the part 
included between these two chords. 

63. The radius of a circle is 7 yards. A chord is drawn 
equal to the side of the regular inscribed polygon of 15 
sides. In the larger of the two segments thus formed, a 
circle is inscribed touching the arc in its middle point. 
Calculate the area of a regular polygon of 9 sides circum- 
scribed about the latter circle. 



CHAPTER V. 
OBLIQUE TRIANGLES. 




35. Notation. — In an obliqne triangle, let A denote the 
value of angle A, B the value of angle B, C the value of 
angle C ; a the length of the side opposite A, b the length 
of the side opposite B, c the length of the side opposite 0; 
p the perpendicular from C on c, c x the distance from A to 
the foot of the perpendicular, and c 2 the distance from B 
to the foot of the perpendicular. The length c x will be 
regarded as positive when drawn from A toward B, and 
negative when drawn in the opposite direction ; c 2 will be 
regarded as positive when drawn from B toward A, and 
negative when drawn in the opposite direction. 

Relation between the angles. — From geometry, 

A.+ B+ (7 = 180°. 
Hence, when any two angles of a triangle are given, the 
third angle can readily be found. 

36. Methods of Solution. — At least three independent 
conditions must be given to make the solution of an oblique 
triangle determinate. Suppose that these three conditions 
are the values of sides or angles. When possible, draw an 
altitude such that two of the given parts are in one triangle. 

66 



METHODS OF SOLUTION. 67 

Now find a value of this altitude in terms of two known 
parts in the one right triangle. Also in the other right 
triangle find a value of this same altitude in terms of the 
other given part and a desired part. These values of 
the altitude give an equation in which the desired part is 
the only unknown. Thus, 

Given one side and two angles : as b, A and B. To find a, 
equate values of p thus : p = a sin B = b sin A. 

Multiply by esc B : a = b esc B sin A. 

If C and c are also desired : C = 180° - (A + B). 
By drawing the altitude from A on a, 

c sin B = b sin C, whence c = b esc B sin C. 
These values of a and c give the following relation : 

acscA = bcscB = c esc C. (19) 

Given two sides and the angle opposite one of these sides: 

as a, b, and A. Equate values of p : a sin B = b sin A ; 

whence sin jB = - sin A, 

a 

C and c being found as in the last case. 

. Given two sides and the included angle : as b, c, and A. 
First find c x = b cos A and c 2 = c — c x . 
To find B, equate values of p thus : 

p = c 2 tan B = Ci tan ^1 ; whence tan B = — tan A 

c 2 
To find a, equate values of p 2 : 

a 2 — c 2 2 = b 2 — c 2 . Put c — Cx for c 2 . 

a 2 = b 2 + (c - C] ) 2 - c 2 = b 2 + c 2 - 2 c Cl . 
Put 6 cos ^1 for Cij then a 2 = b 2 + c 2 - 2 be cos A. (20) 
Given the three sides : as a, b, and c. 

From (20), cos A = 6 * + ** ~ a \ (21) 

with similar relations for the other angles. 



68 



OBLIQUE TEI ANGLES. 



It is usually inconvenient to compute cos A from (21) ; 
another method will now be found for computing the values 

of the angles of a triangle when the three 



nicies are given. 




Fig. 40, 



Prolong the sides AB and AC of triangle ABC. Bisect 
angles A, B. and CBC\. these bisectors meeting at and O v 
The point is the center of the incircle which touches the 
sides of the triangle in A, y B„. and C : the point 1 is the 
center of the excircle in angle A which touches BO at A x . 
AC prolonged at B v and AB prolonged at C v 

From geometry, the tangents from a point outside of a 
circle are equal to each other : thus AB = AQ, AB 1 = AC 1} 
and so on. 



Denote each of the tan- 
gents from J. to the incircle 
by ;/;. each of the tangents 
from B by y. each of the 
tangents from C by z, and 
half of the sum of the sides, 
\ (a — b + c) by s ; then 



Denote each of the tan- 
gents from A to the excircle 
by :i\. each of the tangents 
from B by gfa each of the 
tangents from C by z x . and 
half of the sum of the sides 
\ (a -f h + c) by s ; then 



THE INCIRCLE AND EXCIRCLE. 



69 



y + z = a, 

x + z = b } 

x + y = c, 
which give 

x = ABq = AC = s — a, 
y = BC = BA =s-b, 
z = CA — CB Q = s — c. 



y 1 + z x = a, 
x 1 — z 1 = b, 

Xi — Vi = c, 
which give 

x 1 = AB X = Ad = s, 
y x = BC Y = BA X = s — c, 
Zi = CA X = CB 1 = s-b. 



Now the two triangles AOC and AO x C x are similar be- 
cause their sides are respectively parallel; and the two 
triangles BOC and BO x C x are similar because their sides 
are respectively perpendicular ; hence 



OCo = AC 
Oxd AC, 

OC _BC 



or 



or 



BC X 0& s-c r x 

r and r x being the radii of the incircle and excircle respec- 
tively. 

Multiply these equations and solve for r. 



r Ks-a)(s-b)(s-^c) 
In triangle AOC , 
Changing letters, 



cot J A = 



cot|£ = 



cot^C 



r 
s-b 



r 
s — c 



(22) 



The equations in (22) furnish the solution for the angles 
of a triangle when the sides are given. This method is 
more suitable for logarithmic computation than the use 
of (21). 



70 OBLIQUE TRIANGLES. 



SOLUTIONS. 



One side and two angles. 

37. 1. Given a = 200,. A = 74° 36', B = 81° 12' ; find b. 

By (19), 5 = a cse A sin 5 = 200 x 1.0372 x .9882 = 205. 

2. TVith the same data, find C and c. 

C = 180° - (A + 5) = 180° - 155° 18' = 21° 12'. 
By (19), e = a esc J. sin C=200 x 1.0372 x .4099 = 85. 

3. Given 5 = 37, B = 53° 8', .4 = 107° 57'; find a. 

By (19), a = bescB sin J. = 37 x 1.2500 x .9513 = 44. 

Two sides and the angle opposite one of them. 

1. Given A = 30°, a = 1, 6 = 4 ; find B and C. 
Here sin£ = f sin 30° = 4 x .5 = 2. 
Since sin B > 1, the solution is impossible. 

2. Given A = 30°, a = 2, 6 = 4 ; find B and C. 
Here sin B = § sin 30° = 1. 
Whence £ = 90° and C = 180° -(A + B) = 60°. 

3. Given A = 30°, a = 3, 6 = 4; find B and C. 
Here sin B = a sin 30° = .6667. 

B = 41° 49' or 138° 11', and C = 108° 11' or 11° 49'. 
^Note that the values of C are obtainable by subtracting 
A from the values of B. Explain this geometrically. 

4. Given A = 30°, a = 4L, b == 4 ; find 5 and C. 
Here sin 5 = ± sin 30° = .5. 

£ = 30° or 150°, and C = 120° or 0°. 

5. Given .1 = 30°. a = 5, 6 = 4 ; find £ and C 
Here sin B = 4 sin 30° = .4. 

£ == 23° 35' or 156° 25', and C = 126° 25' or - 6° 25'. 



SOLUTIONS USING NUMBERS. 71 

These five examples are illustrated in the geometric con- 
struction, Fig. 41. The base is drawn indefinitely, angle A 
is made equal to 30°, AC =6 = 4. Successive radii, c^, a 2 , 
a 3 , a 4 , a 5 are used to draw arcs with the center at C. 

In order that the triangles may have the angle A — 30°, 
the vertex B of each of them must be on the half -line AB 5 
drawn through A to the right. Also in order that the tri- 
angles may have the side a of the length given in each 
problem, the vertex B must be in the arc appropriate to 
that problem. 

To prevent confusion in the diagram, the side a has not 
been drawn, except for the negative solution of question 5. 
It is recommended that the boundary of the triangles be 
drawn with crayons of different colors. 




Fig. 41. 

It will be noticed that the first arc does not intersect the 
half -line AB 5 . The second arc touches it at one point. The 
third arc cuts it at two points, both to the right of A. The 
fourth arc cuts it once to the right of A and once at A. 
The fifth arc cuts the half-line AB 5 to the right of A and 
also cuts the half-line AB 5 ' to the left of A; the negative 
value of C = - 6° 25', being the angle ACB 5 \ 



72 



OBLIQUE TRIANGLES. 



Solve the following five examples, and compare the results 
with the geometric construction, Pig. 42. 

1. Given A = 150°, a = l, 6 = 4; find B and C. 
a = 2, 6 = 4; find B and G 
a = 3, 6 = 4; find B and G 
a = 4, 6 = 4; find B and (7. 
a = 5, 6 = 4; find B and G 



2. Given ^ = 150°, 

3. Given ^4 = 150°, 

4. Given ^ = 150°, 

5. Given ^L = 150°, 



«. ": \ 

\ \ 




Fig. 42. 



Assume a value in each of three columns in any row in 
the following table, and compute the remaining value (or 
values) in that row : 



a 


b 


A 


B 


14 


15 


59° 29' 


67° 23' or 112° 37' 


37 


40 


67° 23' 


93° 42' or 86° 18' 


75 


77 


61° 56' 


64° 57' or 115° 3' 


69 


139 


21° 14' 


46° 52' or 133° 8' 


80 


401 


11° 27' 


84° 17' or 95° 43' 


101 


120 


43° 36' 


55° V or 124° 59' 


148 


153 


53° 8' 


55° 48' or 124° 12' 


240 


409 


34° 7' 


72° 56' or 107° 4' 


449 


560 


51° 25' 


77° 10' or 102° 50' 



SOLUTIONS USING NUMBERS. 



73 



6. Given a = 7, b = 8, A = 60° ; find c directly. 
From (20), c 2 - 2 be cos A = a 2 - b 2 . 
Substituting the given numerical values, 

c 2 - 8 c = - 15, 
which gives c = 3 or 5. 

7. Given a = 7, & = 5, ^ = 120° ; find a 
Here the equation is c 2 -f- 5 c = 21. 
Therefore c = 3 or — 8. 

Two sides and the included angle. 

1. Given b = 85, c = 200, A = 81° 12'; find jB. 
ex = 6 cos .4 = 85 cos 81° 12 1 = 13. c 
c 2 = 200-13=187. 
p= c 2 tan B = e x tan A 

tan £ = - tan .4 = . 4491 
Therefore B = 24° 11. 

2. With the same data, find a. 
From (20), a 2 = b 2 + c 2 - 2 6c cos .4 




Therefore 



= 7225 
a = 205. 



40000 - 5200 = 42025. 



3. Given b = 37, c = 15, ^L = 107°57'; find B. 
Here c x = b cos J. = — 11.4. 

c 2 = 15 -(-11.4)= 26.4. 

tan B = ^ tan J. = - ~t* x (- 3.0868) 

C2 ^ O . TC 

= 1.3329. Therefore, B = 53° 7'. 

4. With the same data, find a. 
a 2 = b 2 + c 2 -2bc cos ^L = 1369 + 225 + 342 = 1936. 

Therefore, a = 44. 




74 OBLIQUE TRIANGLES. 

The three sides. 

1. Given a = 7, 6 = 8, c = 5; find A. 

2 be 2x8x5 

Or the angle may be found by using the first and second 
equations in (22) ; thus, 



2. Given a = 7, 6 = 5, c = 3 ; find A 

cos J. = t> 2 + c 2 -a? = - .5. Therefore A = 120°. 

2 6c 



cot 4 J. =\ &5 X ; 5 , = .5774 ; i^=60°. 
*2.5 x 4.o 

Exercises. — Assume any three parts in each of the fol- 
lowing five triangles. Then compute the fourth part : 

^1 = 60° a = 13 6 = 15 c = 7 or 8 

^ = 120° a = 13 6= 7 c = 8or-15 

.4 = 53° 8' a = 13 6 = 15 c = 4orl4 

^. = 61° 56' a = 65 6 = 68 c = 7or57 

^L = 67°23' a = 25 6=26 c = 3 or 17 

38. Given two sides and the included angle, to find a func- 
tion of another angle directly in terms of the given parts. 

Since c = c T -f- c 2 : 

c = p (cot A -f- cot B). 

Multiply by esc A, and put 6 for p esc A : 

c esc .4 = 6 (cot A + cot £). (23) 

Whence cot B = c ^^ - cot A (24) 

b 

After 12 has been found, a = esc I? -= 




BROCAED'S POINTS. 75 

The objection to the use of (24) is that in general both 
tables of functions are used; the table of logarithms of 

functions in the computation of the value of c — , and 

b 

natural values of functions in completing the work. In 
general, this computation is not troublesome ; and where 
results are to be carried through a series of triangles, as in 
a survey, it is of advantage to be 
able to write down the function 
of a desired angle direc'tly in 
terms of the data. 

As a simple application of (23), 

consider the following problem : 

& r Fig. 45. 

When a point (X) in a triangle 

is joined to the corners, it is found that the angles XBA, 

XCB, and XAC are all equal. Find the value of one of 

these equal angles (x). 

In triangle AXC, by (23) ; ^Xcscx=6(cot#+cot CXA). 

In triangle AXB, by (19) ; AX esc x = c esc AXB. 

Now angle CXA = 180° -{C-x)-x = 180° - C. 

Similarly, AXB = 180° - A. 

Hence by substituting : b (cot x — cot C) = c esc A. 

But by (23), c esc A = b (cot A 4- cot B). Hence 

cot x = cot A + cot B + cot C. 

If the angles are 65°, 30°, and 85°, respectively, x = 23° 38 f . 

The point X and another point found by counting the 
angles x from the other sides of a, b, and c are called 
Brocard's Points. They possess many interesting properties; 
for example, they are the foci of the ellipse inscribed in 
the triangle. 



76 OBLIQUE TRIANGLES. 

THE AREA OF A TRIANGLE. 
39. Denote the area by K. Then 
By geometry, K=±cp. \ 

Since p = b sin A, K=\bc sin A. 

Changing letters, K= \ ca sin B. 

K=\ab sin C. 
b = acscAsmB, K= ^a 2 cscAsinB sinC. 

Changing letters, K= -J- b 2 sin A esc B sin C. 

K=±c 2 sin A sin B esc C. 
From Fig. 40, K= | ar 44 &r + i cr = rs. 



Substitute from (22), If = Vs (s - a) (s -b)(s- c). 



Prove the following : K= ±~vd 2 b 2 c 2 sin J. sin B sin (7. 

7f = r 2 cot -i J. cot±B coti C. 

K= r x 2 cot i ^L tan -J- 5 tan f (7. 

iT= r^coti A 

ir= s 2 tan i A tan 1 5 tan \ C. 

K= (s-a) 2 tani^cot|jScotia 

By equating any two values of the area, a relation between 
segments and angles of the triangle is obtained. 

Exercises. — Without using logarithms, find the area of 
each of the following triangles : 

a = 115 £ = 67° 23' C=46°24' 

a = 240 5 = 53 B= 8° 10' 

a = 200 6 = 205 0=24° 11' 

a = 116 b = 105 c = 143 



SOLUTIONS USING LOGARITHMS. 77 

LOGARITHMIC SOLUTIONS. 
40. Given one side and two angles. — 

a = 24.31, 4 = 45° 18', B = 22° 11'. 
Solve the triangle. 

= a esc A sin B, c = acscA sin C. 





Zsin£ 9.5770 




log a 1.3858 


A + B= 67° 29' 


log esc A 0.1483 


C=112°31' 


L sin 9.9656 


b = 12.92 


log b 1.1111 


c= 31.6 


log c 1.4997 



Given two sides and the angle opposite one of them. — 
a = 215.9, b = 307.7, A = 25° 10'. 

Solve the triangle, 
sin B = - sin J., d = -B 2 — A, C 2 = B 1 — A, c = a esc A sin 0. 

log b 2.4881 

log a 2.3343 





0.1538 
XsinJ. 9.6286 


B 1 = 37° 18' 


LsinB 9.7824 


£ 2 = 142°42' 
.4= 25°10 f 
d = 117° 32' 
C 2 = 12° 8' 


isinCi 9.9478 
log a 2.3343 
log esc A 0.3714 
L sin 2 9.3226 



c 1= = 450.3 logcj 2.6535 

c 2 = 109.2 log c 2 2.0383 



78 



OBLIQUE TRIANGLES. 



Given two sides and the included angle. 
6 = 767, c = 947, A = 10° 50'. 
Find the remaining parts. 

d= b cos A, c 2 = c — c lf , 

£> = c 2 tan B = c 1 tan J. j 

whence tan B = — tan A, a = c 2 sec 5. 




c = 947 
Cl = 753.3 
c 2 = 193.7 



10° 50' 
36° 40' 



A 

B 

47° 30' 
0=132° 30' 
a = 241.5 



log b 

LcosA 

logcj 

logc 2 

Lt&nA 
LttmB 
log sec B 
logc 2 
log a 



2.8848 
9.9922 
2.8770 
2.2871 
0.5899 
9.2819 
9.8718 
0.0958 
2.2871 
2.3829 



Given b = 767, c = 947, A = 10° 50' ; find a. 
By (20), a=^b 2 + c 2 -2bccosA. 



b 2 = 588300 



896600 



r>2_ 



log 2 
log b 
logc 
L cos ^4 
log& 2 



2 6c cos A 



1484900 
1427000 



0.3010 
2.8848 
2.9763 
9.9922 
5.7696 
5.9526 



a 2 = 57900 
a = 240.6 



logc 2 

log 2 be cos A 6.1543 

log a 2 4.7627 

log a 2.3813 



Two significant figures were lost in the subtraction of 
2 be cos A from b 2 + c 2 . Hence the lack of precision in the 
value of a. 



SOLUTIONS USING LOGARITHMS. 



79 



Given h = .049, c = .045, A = 143° ; 
find the remaining parts. 

c L =h cos A, c 2 = e — Cu 



tan B- — tan A 
a = c 2 sec 5. 

c = .045 


Lh 
LcosA 

Lci 

Lc 2 

Lt&riA 
Lt&nB 

log sec B 

Lc 2 

La 


P \ X 
c< \A c \ 

Fig. 47. 

8.6902 
9.9023 n 


Cl = - .03913 


8.5925 ?i 


c 2 = .08413 


8.9250 


A = 143° 


9.6675 ?i 
9.8771 n 


£ = 19° 19' 


9.5446 


162° 19' 
C= 17° 41' 


0.0252 
8.9250 


a = .08917 


8.9502 



Given h = .049, c = .045, ^L = 143° ; find a. 



a = VF 


+ c 2 


— 2 he cos 

log 2 
Lh 
Lc 
LcosA 

Lh 2 


A. 


0.3010 
8.6902 
8.6532 
9.9023 n 


.002401 


7.3804 


.002025 




Lc 2 




7.3064 


.003521 




L 2 be cot 
La 2 


A 


7.5467w 


.007947 


7.9002 


.08915 




La 




8.9501 



80 



OBLIQUE TRIANGLES. 



Given the three sides, — a = 704, b = 302. c = 670. 
Find the angles and the area. 



j(s-a)(s- 
s 




cot \A = — 
2 r 


— , etc. ; 


704 


134 


logs — a 


2.1271 


302 


536 


log s — b 


2.7292 


670 


168 


log s — c 


2.2253 


1676 


7.0816 


868 


868 


logs 
logr 2 


2.9232 




4.1584 






logr 


2.0792 


83° 42' 




log cot \ A 


0.0479 


25° 14' 




log cot \B 


0.6500 


71° 4' 




log cot \ G 


0.1461 


100,550 




log if 


5.0024 



K=r 



Exercises. — Assume three values in any row of the follow- 
ing table and compute the remaining values in that row. 



a 


b 


c 


A 


B 


K 


6.82 


5.20 


3.16 


106° 47' 


46° 53' 


7.864 


.317 


.533 


.510 


35° 18' 


76° 19' 


.07854 


28.9 


60.1 


71.2 


23° 32' 


56° 9' 


854.3 


1.98 


2.02 


3.16 


37° 22' 


38° 16' 


1.937 


312 


109 


229 


131° 25' 


15° 11' 


9360 


14.8 


17.5 


15.3 


53° 8' 


71° 5' 


107.1 


1.75 


1.62 


1.19 


67° 23' 


73° 44' 


99.96 


145 


119 


156 


61° 56' 


46° 24' 


8190 


11.6 


40.4 


48.0 


11° 25' 


43° 36' 


192.0 



SOLUTIONS OF PROBLEMS. 



81 



41. Whenever it is possible, by means of the ruler and 
compasses, to obtain the geometric construction of a prob- 
lem, the trigonometric solution is always obtainable by follow- 
ing the order of the construction. It occasionally happens, 
however, that the geometric construction is not possible. 
When this occurs, the trigonometric solution produces an 
equation of a higher degree than the second. In many such 
cases, the method of solution by trial involves the least labor. 

As an illustration, consider the following problem : 

A ship is sailing in the direc- 
tion N. 35° W. When at B, two 
lighthouses are sighted, one due 
north, the other due west. After 
sailing 3 miles, the ship was 
equally distant from the light- 
houses ; sailing 1 mile further 
on the same course, the ship and 
the lighthouses are in the same 
straight line. Find AC, the dis- 
tance between the lights. 

From D, the second position 
of the ship, drop a perpendicular 
to F, the middle of AC. Denote angle A by x ; then the 
geometry of the figure gives: angle BFE = 2 x, BEF = 
125° - x, BFD = 2x- 90°, and BDF = 215° - x. 

Apply (19) to triangles BFD and BFE : 
BF= 3 esc (2x- 90°) sin (215° - x) = 4 esc 2 x sin (125° - x), 
or 3 sec 2 x cos (125° - x) = 4 esc 2 x sin (125° - x). 

Multiply by sin 2 x sec (125°— x) and transpose : 
3 tan 2 a; -4 tan (125° -a) = 0. 

Now solve this by trial. Substituting 0° for x makes the 
left-hand member of the equation positive, 90° — , 45° -+-, 
60°-, 50°+, 52°-, and so on; x = 5l°V. Hence 
AC = 2 BF= 8 esc 102° 14' sin 73° 53' = 7.864 miles. 




Fig. 48. 



82 OBLIQUE TRIANGLES. 

Iii order to measure the dis- p | j 

tance between two inaccessible qJ\ /\d 

points, C and D, a base line » \ / 

JLjB was laid off equal to 583 ^J^^^m^d^^^^ 

feet (a). At J. angle CAD " —E^r^^^^ 

= 34° and angle DAB = 47° ,' / \ \ 

were measured. Also at B, the ' / \ \ 

angle JJ3C = 44° and angle {/ V 

CBD = 32°. Find CD. A "" „ 4n B 

Fig. 49. 

AC=acscACB sin ABC; AD = acscADB sin ABD] 

CD 2 = AC 2 + AD 2 - 2 AC x AD cos CAD J (2d) 

Z sin .450 9.8418 



„40B = 


= 55° 


log esc AC B 


0.0866 






log a 


2.7657 


.41)5 = 


= 57° 


log esc ADB 


0.0764 


ABD = 


= 76° 


L sin ABD 


9.9869 






log AC 


2.6941 


AC 2 -- 


= 244500 


log AD 


2.8290 


AD 2 -- 


= 455000 


log 2 


0.3010 




699500 


Z, cos CAD 


9.9186 




553000 




5.7427 






log ^4C 2 


5.3882 






log AD 2 


5.6580 


CD 2 -- 


= 146500 


log CD 2 


5A658 


CD-- 


= 382.7 


log CD 


2.5829 



In making an extended survey, it may happen that CD, 
Fig. 49, is a base of known length, the angles being meas- 
ured from A and B. In this case, to find AB, assume the 
figure drawn to scale, AB being represented by 1 foot. In 
this reduced figure, CD may be computed ; then the numeri- 
cal value of AB equals the ratio of the known length of CD 
to the computed length of CD in the reduced figure. 



SOLUTIONS OF PROBLEMS. 



83 



The sides of a trian- 
gle are : a = 57, b = 51, 
c = 64. A point is taken 
in the side c, such that 
the segments of c are : 
h = 21, k = 4:3. Find I, 
the distance of this point 
from C. 

Denote the angle be- 
tween h and I by D. 

By (20), I 2 + h 2 - 

By (20), l 2 + k 2 -2kl cos (180° - D) = a 2 . 

Multiply the first by &, the second by h, and add 

cZ 2 -f chic = aVi + b 2 k. 
Substituting, I is found to be 43.71. 




2hlcosD = b 



— 7.2. 



(26) 



The top of a lighthouse, known to be 100 feet (h) high, 
is just seen in the horizon. What is the distance (6) of the 
lighthouse from the observer ? 

It is known that near the earth's surface light does not 
move in a horizontal line ; on account of refraction, a ray, 
which seems horizontal, moves in a curve, which is nearly 
the arc of a circle whose radius is 7 times the earth's radius. 

In (26) put R for h, 6 R for k, 7 R for a or c, R + h for I 
Then 6b 2 = 7(2Rh + li 2 ). 

In the problem under consideration h 2 is very small com- 
pared to 2 i?/i, and may be neglected. Now h is usually 
given in feet, and b is desired in miles. Neglect h 2 and 
multiply the left side of the equation by 5280 : 

31680 b 2 (miles) = 14 R x (h in feet). 

Put 3960 for R and solve for b. 



b (in miles) = 1.323 Vh (in feet). 
Substituting, b — 13.23 miles. 



(27) 



84 



OBLIQUE TRIANGLES. 




Fig. 51. 



Snell's Problem. — On a coast are three prominent objects, 
A, O, and B. CB = 1100 feet (a), AC = 1800 feet (b), and 
the angle ACB = 150° (C). A buoy is placed at a point 
X, and from a boat at the buoy the horizontal angles 
AXC = 80° (D) and CXB = 60° (J2) are measured. Find 
the distance from X to C* 

The following values are obtained by following the geo- 
metric construction : angle AOC = 2 D (same arc), 
angle BO x C =2E, angle OC0 1 =C + D + E- 180°. 

Hence OC — \a esc D, and ± = i ft esc E. 

Now denote angle C00 1 = XAC by A. From (24), 



cot^=— -cscDsin^csc(C+D+^)-cot(C+D+ J E7) 
OX = a esc D sin ^4 



K28) 



* By many writers the credit has been given to Pothenot for the 
suggestion of this method of determining the position of a point by 
means of angles subtended by segments joining three known points. 
The problem, however, was proposed and solved in the year 1617 by 
Willebrord Snell, the discoverer of the law of the refraction of light. 
John Collins gave a solution of the problem in the Philosophical 
Transactions for 1671 : Pothenot's solution was in 1692. 



SNELL'S PROBLEM. 



85 



terms of the data, CX 



area OC0 1 



00 1 

ab esc D esc E 



2 Va 2 cse 2 D + b 2 esc 2 E + 2 ab esc Z> esc # cos (0 + Z> 

The computation by (28) proceeds thus : 
log ( - a) 3.0414 ?i .5719 log a esc Z> 3.0480 

log esc D 0.0066 nat cot 290° - .3640 
LsuiE 9.9375 nat cot A .9359 
log esc 290° 0.0270 n A = 46° 54 ' 



E) 



isinJ. 9.8634 

log CX 2.9114 

OX = 815.5 



log& 



3.0125 
3.2553 

9.7572 




In laying out a railroad 
along the seashore, the dis- 
tances CB and BE were 
measured (C, B, D, and E 
all in a straight line) ; also 
the angles CAB, BAD, and 
DAE were measured. It 
was found that CB = 2200 
(a), DE = 2300 (b), CAB = 
21° 30' (A), £-4# = 20° 26' 
(A,), and DAE = 22° 4Q' 
(A 3 ) ; find £Z>. Fig. 52. 

Denote J3Z> by x. Then 

In triangle ADC, x + a = AC esc D sin (^ -f- A 2 ). 

In triangle J..EJ.B, « + b = AE esc B sin ( J. 2 -f ^4 3 ). 

In triangle ABC, AC = a esc A x sin 5. 

In triangle AED, AE = b esc A 3 sin Z). 
The product of these four equations is - 
(x+a)(x+b)=ab csc^csc^sin^+^sin^-fA)- (29) 
This quadratic equation gives x = 1800. 




86 



OBLIQUE TRIANGLES. 



There are two observatories on the same meridian : A in 
Sweden, latitude 65° 30' K (I) ; B at the Cape of Good Hope, 
Z -— P 




Fig. 53. 



latitude 34° 20' S. (— I). When the moon passed this merid- 
ian, the angles ZAP = 62° 20' (z) and Z^P = 39° 20' (z x ) 
were measured. The radius of the earth being 3960 miles, 
find the distance from the moon to the earth's center. 

The angles of the quadrilateral AEBP are readily found : 
at E, the angle is I — l x ; at A, it is 180° — z ; at P, it is 
z + z 1 -l + l 1 . Also the angle ABP is 90° + \ (I -h - 2 %). 
In triangle AEB, AB = 2AE sin 1 A#£. 
In triangle ^4£P, ^4P = .45 esc ^LPP sin ABP. 

Denote AE by r, AP by v, i?P by x, angle ^IjEJP by w. 



v=2r sini(Z— 1{) esc (z+z{— l+ty cos-] 


r (7-^-2*0 




cot w = 


r 

-- - esc z -f cot z 

V 




(30) 


x = 


-- v csc w sin ^ 






log 2 0.3010 


£- 8.2257 


logv 5.3720 


logr 3.5977 
I, sin 49° 55' 9.8837 
log esc 1°30' 1.6332 
Lcos 25° 15' 9.9564 
logv 5.3720 


log csc z 0.0122 

8.2379 

.0173 

nat cot z .2401 

nat cot iv .2574 


log csc w 

L sin z 9 

5 

a* = 236,40 


.0139 
.9878 
.3737 
Omi. 



APPLICATIONS. 87 



PROBLEMS. 



42. 64. Solve Problem 10 by logarithms. 

65. Solve Problem 11 by logarithms. 

66. Solve Problem 12 by logarithms. 

67. Solve Problem 13 by logarithms. 

68. Solve Problem 14 by logarithms. 

69. Solve Problem 15 by logarithms. 

70. To measure the distance between two points A and C 
(C inaccessible), a base line AB = 693 feet was measured 
from A. At A, the angle subtended by BC was 68° 29' ; 
and at B, the angle subtended by AC was 66° 7'. Find the 
distance AC. 

71. To find the distance AB across a pond, a third point 
C was taken. AC =735 yards, BC = 810 yards, and the 
angle ACB = 55° 40' were measured. Find AB. 

72. From a window on a level with the bottom of a 
steeple, the angle of elevation of the' top is 40°. From 
another window 18 feet directly above the former the 
angle of elevation is 37° 30'. Required the height of the 
steeple. 

73. A tower 50 feet high is situated on a cliff over the 
sea. From the bottom of the tower, the angle of depres- 
sion of a ship at anchor is found to be 25° 17' ; and from 
the top the angle is 31° 12'. Find the height of the cliff 
and the direct distance of the ship from the top of the 
tower. 

74. A frigate 10 miles S.W. of a harbor sees a ship sail 
from it in the direction S. 70° E. at the rate of 9 miles 
an hour. In what direction and at what rate must the 
frigate sail in order to come up with the ship in two 
hours ? 



88 OBLIQUE TRIANGLES. 

75. A person is at the bank of a river directly opposite 
a tower on the other side. Going backward from the bank, 
up a slope making an angle of 42° to the horizontal, he 
measures an oblique distance of 528 feet. He there finds 
the angle of depression of the top of the tower is 90°, and 
of the base is 27°. Required the height of the tower and 
the breadth of the river. 

76. To measure the breadth of a stream, a base AB was 
chosen = 490 yards long, parallel to the bank and 50 yards 
from it. A tree C stands on the other bank of the river. 
The angle CAB was found to be 62° 37' and CBA 40° 28'. 
What was the breadth of the stream ? 

77. At the foot of a mountain the elevation of the sum- 
mit = 32° 57'. A person measures 775.3 feet along the slope 
toward the summit, making an angle of 7° 15' with the hori- 
zontal plane ; at this point, the elevation of the summit was 
60° 32'. Find the height of the mountain. 

78. A tower stands on a hill whose slope is uniformly 
26° 37'. An observer at the foot of the hill finds that the 
tower subtends an angle of 10° 22'. Measuring horizontally 
up the slope a distance of 200 feet, he observes the tower 
subtends an angle of 20° 30'. Find the height of the tower. 

79. A tower is situated on a hill. At a point in the hori- 
zontal plane through the foot of the hill, the angle of eleva- 
tion of the top of the hill is 40° and of the top of the tower 
is 51°. Measuring in a line directly away from the hill 
a distance of 180 feet, the angle to the top of the tower is 
found to be 33° 45'. Find the height of the tower. 

80. From a hill 200 feet above the sea, the top of a mast, 
known to be 150 feet above water, was just seen in the 
horizon. How far distant is it ? 

81. A body is thrown with a velocity of 25 feet per 
second horizontally from the window of a railway carriage 
moving at the rate of 30 miles an hour; the direction in 



APPLICATIONS. 89 

which the body is thrown makes an angle of 30° with the 
rear of the train. Find the direction of the vertical plane 
in which the body moves. 

82. A man traveling at the rate of 6 miles an hour on 
a road that went due east observed that the wind struck 
him from the northeast. But having occasion to stop, he 
found that it actually came from the direction N. 35° E. 
Find the velocity of the wind. 

83. A ship's apparent course is N. 33° 45' E., 8 knots an 
hour ; but the tide sets her S. 67° 30' E. at the rate of 3 knots 
an hour. What is her true course and rate of progress ? 

84. Two forces, P — 6.2 lbs. and Q = 10.9 lbs., making an 
angle of 112° 4', act at the same point on a body. Find the 
magnitude of the resultant and the angle which it makes 
with the force P. 

85. Three forces in a plane acting on a body produce 
equilibrium. The greatest is 5 lbs., the smallest 3 lbs. 
The angle between the 3 lb. force and the unknown force 
is 79° 43'. Find the third force. 

86. On the same meridian are two places whose difference 
of latitude is 68° 13'. From these places, the zenith dis- 
tances of the moon at its culmination were found to be 
25° and 44° 18', respectively. The radius of the earth being 
3960 miles, find the distance of the moon from the earth's 
center. 

87. In measuring a distance AD, a portion of it (BC) 
between A and D is inaccessible. AB = 244, CD — 520. 
From a point E outside of AD, angles were measured : 
AEB = 82° 8', BEC= 21° 2', CED = 40° 14'. Find BC. 

88. A man walking along a straight road observes the 
angle of elevation of a tower to be 28° 40'. At a point 
57 feet further on, the angle of elevation of the same tower 
is 48° 42'. And 143 feet beyond that, the angle is 33° 10'. 
Find the height of the tower. 



90 OBLIQUE TRIANGLES. 

89. From a ship, a rock is seen in the direction 1ST. 11° 48' E. 
The ship sails due southeast 5 miles when the direction of 
the rock is 1ST. 1° 36' E. Find the distance from the rock to 
each position of the ship. 

90. The sides of a triangle are three consecutive whole 
nnmbers ; and the greatest angle is double the least (A). 
Find the average side (c). 

91. The distance between the centers of two circles is 
238 (c). The outer common tangents make with each other 
an angle of 36° 8' (A). The inner common tangents make 
with each other an angle of 104° 12' (E). Find the length 
of the radius of the larger circle. 

92. The distances between three points A, B, are as 
follows: AC =1200 feet, £0=1320 feet, and AB = 1±Q0 
feet. An observer at X in the plane ABC finds angle 
AXC = 25° 32'and 5X0 = 40°. Find the distances of X 
from the points A, B, and O, if O and X are on different 
sides of AB and angle AXB = AXC + CXB. 

93. From the top of a cliff 300 feet high, the altitude of 
the sun was observed to be 70°, the angle of elevation of a 
balloon was 52° 12', the angle of depression of its shadow 
on the sea was 6S° 14' The sun is in front of the observer, 
and the cliff, sun, balloon, and shadoAv are all in the same 
vertical plane. Find the height of the balloon above the 
sea. 

94. From a bluff 100 feet above the surface of a lake, the 
angle of elevation of a balloon is 48° 20', and the angle of 
depression of its image reflected from the surface of the 
water is 39° 50'. Find the height of the balloon above the 
lake. 

95. From the deck of a ship sailing due east, two con- 
spicuous headlands are observed to bear 63° and 24°, 
respectively, to the northward of the ship's course. After 



APPLICATIONS. 91 

sailing 8 miles, the corresponding angles were found to be 
150° and 98°. Find the distance of the headlands from one 
another. 

96. Calculate the distance of two inaccessible points A 
and B, knowing a base CD = 600 feet, the angle BCD = 40°, 
ACD = 69°, ADC = 38° 30', and the angle BDC= 70° 30'. 

97. Three points A, C, and B being given on a chart 
of a coast, it is desired to determine the position of a 
fourth point D. AC =1000 feet, 5(7=850 feet, angle 
ACB =114° 40', angle ADC =46° 17', angle CDB = 30° 9'. 
Find CD. [CD is between AD and BD; C and D are on 
opposite sides of AB.~\ 

98. From the top of a lighthouse 103.5 feet above the 
sea, the horizontal angle between two boats was found to 
be 41° 20', and their angles of depression were 13° 23' and 
17° 41' respectively. Find the distance between the boats. 

99. A balloon is observed from two stations 1 mile apart. 
At the first station, the 
horizontal angle of the ^i 
balloon and the other ^'' 
station = 82°, and the ,**' 
elevation of the balloon /'' / 
= 12°, At the second ^ 
station, the horizontal an- „-"' / 



C 



B 



gle between the first sta- A^ s 

tion and the balloon = 65°. \ / /'' 

Find the height of the \ / /' 

balloon. \ / /' 

100. To determine the ^J'' 

height BC of a tower, two 
points A and D 90 feet 

apart were taken in the horizontal plane through its base. 
The angle BAD = 56° 33', BAC=<55° 23', BDC=56° 26'. 
Find BC 



92 OBLIQUE TRIANGLES. 

101. To measure the breadth of a river AD, a point B 
was taken in AD prolonged. From B a base line BC = 400 
feet was laid off, making an angle of 95° 16' with AB. At 
C, the angles BCA = 52° 48' and BCD = 24° 39' were meas- 
ured. Find AD. 

102. From three points A, D, B in a horizontal straight 
line, AD = 500 feet, DB = 300 feet, the angles of elevation 
of a tower were measured, and found to be 9° 30', 51° 14', 
and 27° 51', respectively. Find the height of the tower. 

103. Three points A, D, B lie in a straight line. AD =232, 
DB =80. To locate a fourth point C in the horizontal 
plane of these three, the angles ACD = 85° 12' and DCB = 
46° 13' were measured. Find CA, CD, CB. 

104. The pedestal of a monument is 9 feet high, the 
column is 8 feet high, and the statue is 10 feet high. A 
person is in such a position that he sees the three parts 
subtending the same angle. What angle with the monu- 
ment is made by the line joining his eye and the base of the 
monument ; and what is the height of his eye above the base ? 

105. Within an angle A = 73° 44', a point is chosen whose 
distances from the sides of the angle are 6 inches and 8 
inches. A line is drawn through this point so that the 
portion included between the sides of the angle is bisected 
at the point. Find the area of the triangle thus formed. 

106. The two non-parallel sides of a trapezoid are 7.5 
feet and 6.3 feet. The angle formed by their continuations 
= 41° 21'. The base of the trapezoid is 10 feet. Find its 
area. 

107. From two vertices of an equilateral triangle, whose 
side is 234 feet, two bodies begin to move simultaneously 
toward the third vertex ; one with a velocity of 4 feet per 
second, the other with a velocity of 3 feet per second. When 
will the distance between the moving points be equal to the 
altitude of the triangle ? 



APPLICATIONS. 



93 




Fig. 55. 



108. Calculate the radius of an inaccessible cylindrical 
tower from the following 

data : A base line AB is 
measured = 70 feet. The 
angles formed with the 
base line by the pair of 
tangents are : From A, 
60° and 20°; and from 
B, 75° and 25°. 

109. At some distance 
from a circular pond, a 
base line AB = 250 feet 
is measured. At A the 
angles between the base 

line and the tangents to the pond are 71° 12' and 54° 12'. 
At B, the angle between the base line and the tangent to 
the right-hand side of the pond is 57°. Find the radius 
of the pond. 

110. At some distance from a circular pond, a base line 
AB = 75 yards is measured. At A the angles made by the 
base line and the tangents to the pond are 72° 58' and 44° 2'. 
At B the angle between the base line and the tangent to 
the left side of the pond is 38° 20'. Find the area of the 
pond. 

111. To find the height of a hill, three points, A, B, and C, 
not in a straight line, were found, such that at each of the 
three points the angle of elevation of the summit of the hill 
was the same ; viz. 52° 40'. CB was measured, and found 
to be 121.6 feet, and angle BAG = 15° 25'. Find the height 
of the hill. 



CHAPTER VI. 



PROPERTIES OF TRIANGLES. 
C 




43. The altitudes. — p = a sin B = b sin A. Similar ex- 
pressions hold for the other altitudes. 

If the three altitudes are known, the triangle can be 
solved. Denote the altitudes by p x , p 2 , and p 3 . Then 



2K 

a = f 

Pi 

Substitute in (21) ; 
1 



b = 



2K 

i>2 ' 



C = 



2K 

Ps 



cos^l = 



p* pi 



Pi 



PzPi+PiPi 



pipi 



2 
P2P3 



2pi 2 P 2 Ps 



with similar expressions for cos B and cos C. Then 
b = 2h esc A-, c=p 2 esc A, etc. 

The bisectors of the angles. — Let CQ (= q) be the bisector 
of the angle ACB. Then 

area BCQ + area QCA = area BOA ; 

\ aq sin \ C + | bq sin \ G = \ ah sin C = ab sin ^ C cos |- C. 

94 



ALTITUDES, BISECTORS, AND MEDIANS. 95 

W i 2 ab cos 4- C 

Whence q = 2 — 

a + b 

Now angle PCQ = \{A — B)\ hence q = p sec J (A — .B). 

Hence g = a sin 5 sec %(A — E). 

Similar values can be obtained for the other bisectors. 

AQ = b esc AQC sin \ C; QB = a esc CQB sin ± C. 

Since esc ^QC = esc CQB, 4^ = ~ 

QB a 

But AQ+QB=c; hence .4Q = — — and Q^ = 



a -\- b a + 5 

The medians. — Let 3/ be the middle point of AB. By (20), 

a 2 = i c 2 + m 2 — c??i cos BMC-, 

b 2 = \ c 2 + m 2 — cm cos AMC. 
Now cos BMC + cos AMC = 0. Add and solve for m. 



m = |V2a 2 + 2 5 2 -c 2 . 

Similar relations hold for the other medians. 

If the angles are known and the medians are desired, find 
the angle at M first. Denote the middle point of BC by 
M XJ the middle point of CA by M 2 . Then by (24), 

cot AMC = ccscA - cot A 
2b 

= J (cot B -cot A). 
Similarly, cot BM X A = \ (cot C - cot B). 
cot CM 2 B = J (cot A - cot C). 
Check : cot AMC + cot £JV1 + cot <m>5 = 0. 
If the three medians are known, m x , m. 2 , m 3 , the sides can 
be found. Their values can be obtained from these equa- 
tions : 

9 a 2 = 8 mi + 8 m 2 - 4 mi, 

9 6 2 = 8m 3 2 + 8m 1 2 -4m 2 2 , 
9 c 2 = 8 To! 2 + 8 mi - 4 mi. 



96 



PROPERTIES OF TRIANGLES. 




Fig. 57. 



44. The incircle and the excircles. — Let be the center 
of the incircle touching the sides of the triangle ABC in A , 
B , and C ; lf the center of the excircle in angle A and A ly 
B x , Ci, its points of tan gen cy, etc. Then 

AG X = AB X = BA 2 = BC 2 = CB S = CA Z = s, 
AC = AB = BA S = BC 3 = CB 2 = CA 2 = s-a, 
AC 3 = AB 3 = BAs = BC = CB 1 =CA 1 = s-b, 
AC 2 = AB 2 = BA X = BC X = CB =CA = s-c. 

Denote the radius of the incircle by r, the radii of the 
excircles by r x , r 2 , and r 3 respectively. 

Values of the various segments in the figure will now be 
obtained in terms of the sides. 

OC' = OB sin 15; OB = a sin \ C esc BOC(= 90° + £ A), 
r = a sec \ A sin 1 B sin i C 
= b sin i A sec ^ B sin i 
= c sin i ^4 sin 1 5 sec ^ C. 



THE INCIRCLE AND EXCIRCLES. 97 

Since 0^= O^cos^B, 

and X B = a esc CO x B(= 90° -\A) sin 1 CB(= 90° - |0). 

Also, 0,B = c esc ^L0 1 J5(= i C) sin ± A. 

r x = a-sec i ^ cos ^ 5 cos ^ C 
= b sin i ^4 esc J B cos i (7 
= c sin i J. cos 1 5 esc i (7. 

Find 2 C 2 (= f 2 ) and 3 C 3 (= r 3 ) in a similar manner. 

Since AC X = i\ cot \ A. 

s = a esc i A cos .J JB cos i (7 
= 6 cos i ^4 esc | i? cos i (7 
= c cos \ A cos ^ B esc |- (7. 

^(7 = rcotii4; BC Q = rcot±B; CAo=rcot±a 

s — a = a esc | ^4 sin i J5 sin 1 C 
= b cos i A sec J J3 sin | C 
= c cos \ A sin i J5 sec ^ (7. 

s — b = a sec i ^1 cos -^ -B sin i C T 
=p 6 sin i ^4 esc J^ B sin i (7 
= c sin i ^4 cos \ B sec i (7. 

s — c = a sec |- ^ sin i JB cos J C 
= b sin i A sec ^ 5 cos \ C 
= c sin i J. sin -J- JB esc | (7. 
^0=(5-a)sec|^4; J30=(s-6)sec-J-5; <70=(s-c)sec i (7. 
A0 1 = ssec^A] BOi=(s — c) csc ±B; CO i =(s — b) esc \ C. 
A0 2 =(s—c) esc ^ A; B0 2 =s sec \B; C0 2 =(s— a) esc ^ C. 
AO s —(s — b) esc i A; B0. 3 =(s— a) esc ^J3; <70 3 =s sec i (7. 
OO^^Ox-^lO^a sec i^l; 00 2 =& sec^JB; 00 3 =c sec |(J. 
2 3 =--A0 2 -\-A0 3 — a esc ^ A; O s 1 = bcsc^B-, O x 2 =ccsc\C. 
Angle 
O 3 O 1 O 2 =90°-i^; O 1 O 2 O 3 = 90°-i5; OAO^ 90° -i<7. 



98 



PROPERTIES OP TRIANGLES. 



Functions of the half-angles of a triangle. 



sin 2 \A = 



tan 2 iyl — 



_ ( S -b)(s-c) 
be 
bVs-c) 



2 1 



sec 



2 ±A 



be 



a) 



s (.9 — a) 



esc 



cot 2 \A 



COS 2 A J. = 





6c 




(«- 


-&)(.- 


«) 


i 


•(s-a) 




(s- 


-&)(•- 


«) 


s(s 


— a) # 





6c 



With similar relations for l J3 and -J- C 



45. The circumcircle. — Let 

F be the center of the cir- 
cumcircle, and M the middle 
point of AB. Then since 
angle AI'B at the center of 
the circle is twice the angle 
C on the circumference, 
angle AFM= C. 

Now AF = AM esc AFM. A' 1 
Therefore, changing letters, 



B — la csc ^4 

= \ e esc C. 



\ b esc 5 




]STow K=\be sin ^4 = rs. Hence i? = 



Also, R sin A = ± a, RsinB = ±b, R sin (7 = \ c ; 
the sum and product of which give 

i?(sin A -}- sin B + sin (7) = s, 
R 3 sin ^4 sin B sin (7 = £ a&c = J i£A". 

46. Relations between the sum or difference of two sides and 
the sum or difference of two angles. 

Let ABC be the triangle under consideration. With C as 
a center and radius equal to CA (5), describe a semicircle, 
cutting BC in D and E. Then BD = a-b and BE = a + b. 



DIFFERENCE OF TWO ANGLES. 




Angle DAE = 90° (in semicircle), angle E (on circumfer- 
ence) = i C, angle BDA (exterior) = 90° + \ C, angle DAB 
= 180° - B - (90° + 1 <7)= \{A - 5). 
In triangle ABD, by (19), .4J3 sin DAB = DB sin .4.DB. 

c sin ±-(4 - £)=(« - b) cos i C. (31) 

In triangle AEB, by (19), AB sin ^5 =^7.B sin E. 

c cos l (J. - B) = (a + b) sin i 0. (32) 

The quotient of (31) and (32) reduces to 

(a + b) tani(^ - J5) = (a - b) tan \{A + B). 
The product and difference of the squares reduce to 
c 2 sin [A-B) = (a 2 - b 2 ) sin C. 
c 2 cos(A-B)=2ab- (a 2 + b 2 ) cos C. 
Eliminate a 2 by (20), and a and b by (19). 

sin (^i - J3) = sin C-2 sin 5 cos A, (33) 

cos (J. - B) = 2 sin .4 sin B - cos <7. (34) 

The other functions of A — B can then be found. Since 

A + B = 180° — C, any function of A + J5 can be expressed 

in terms of functions of C. 

Also, since B + 1(7= 90° -\{A - B), any function of 
B -\- ±C is equal to the cof unction of \ {A — B). 



100 PROPERTIES OF TRIANGLES. 

47. General method of solving a triangle, given three seg- 
ments in or about the triangle. — Suppose that to solve the 
triangle there is given any three segments (i?, r, i\, s, s — a. 
etc.). Each of these segments may be expressed in terms 
of the sides and functions of the half angles ; these func- 
tions may be expressed in terms of the sides. Hence the 
sides may be found by solving the set of equations thus 
obtained. 

Relations between the segments. — By using values of any 
two of these segments a relation between the segments may 
be obtained. For example, find r and 5 in terms of R. 
r = a sec \ A sin \ B sin \ C. 
a = 2 R sin A = 4 R sin \ A cos \ A. 
Hence r = 4 22 sin \ A sin \ B sin \ C. 

Similarly, s = 4 R cos \ A cos \ B cos J C. 

48. Points of intersection of the circumcircle. — Let tri- 
angle ABC, Fig. 60, be the triangle under consideration, the 
bisectors of the interior and exterior angles being drawn as 
in Fig. 57. 

Also, let F be the center of the circumcircle of the tri- 
angle ABC. Since angle M'"CO is a right angle, it is in- 
scribed in a semicircle ; hence 3I'"F passes through K'", 
the intersection of the circumcircle and C0 3 . The lengths 
of 2 M ,,f and of O s K '" will now be determined. 
Since FCB = 90° - A (see Fig. 58), 

angle FCK 1 " = CK"'F = \Q- (90° - .4) = \(A - B). 

CM'" = 2Ksul±(A- B )> CK "' = 2 R cos U A ~ B )- 

But R = i c sec \ C esc % C, 

c sin i(^4 - B) = (a - b) cos I C, 
c cos | (J. — B) = (a + 6) sin \ C. 
CM'" =i(a - 6) esc i C; CK'" = }(a + 5) sec i C. 
ajr" = ic csc| C; 3 /r" = ±c sec^O. 



INTERSECTIONS OF THE CIRCUMCIRCLE. 



101 



Whence M'" and K'" are the middle points of X 2 and 00 3 , 
respectively. 

In the same manner it may be shown that M\ K', M", K", 
are the middle points of 2 3 , 00 } , Oi0 3 , 00 2 , respectively. 




Fig. 60. 



If OC (=r) is perpendicular to AB, the angle C OK'" is 
also equal to \{A — B), and, therefore, K'"F is parallel to 
OC ; it is, therefore, perpendicular to AB through its 
middle point M. Hence, K'" is the middle point of the 



102 



PROPERTIES OF TRIANGLES. 



arc AB, K' is the middle point of the arc BC, and K" is 
the middle point of the arc CA. 

The circumcircle of ABC is called the nine-points circle of 

the triangle O x 2 3 . 




49. The nine-points circle. — Let ABC be the triangle 
under consideration. Draw the altitudes AP lf BP 2 , CP 3 , 
meeting in H. Join P X P 2 , P^Pz, P&P\- 

In the triangle AP 3 C, AP 3 = b cos A, 
and in triangle AP 2 B, AP 2 = c cos A. 

Therefore the triangles AP 2 P 3 and ABC having a com- 
mon angle and the including sides in each triangle propor- 
tional are similar ; hence P 2 P 3 = a cos A. 

In a similar manner P 3 P X = bcosB and P X P 2 = c cos C. 



THE NINE-POINTS CIRCLE, 103 

Also angle BP y P 3 = CP,P 2 = A, whence P^P, = 
180° -2 A, PiPtPs = 180° - 2 B, P 2 P S P 1 = 180° - 2 O. 

The intersection .H" Of the altitudes is called the ortho- 
center of the triangle ABC; the triangle P X P 2 P^ joining 
the feet of the altitudes, is called the pedal triangle of the 
triangle ABC. It should be noticed that the orthocenter 
of a triangle is the center of the incircle of the pedal 
triangle. The circumcircle of the pedal triangle is the nine- 
points circle of the original triangle. 

All of the properties of any triangle ABC can be reduced 
to properties of its pedal triangle P X P 2 P S by substituting 
in any formula referring to the triangle ABC, 180° — 2 A 
wherever A occurs, 180° — 2B wherever B occurs, 180° — 
2 C wherever C occurs. After completing these substitu- 
tions, put a cos A wherever a occurs, b cos B wherever b 
occurs, and c cos C wherever c occurs. For instance, make 
these substitutions in the value of P. The radius of the 
circumcircle of P X P 2 P 3 or the nine-points circle of ABC 
= \ a cos A sec \ (180° - 2 A) esc J (180° - 2 A) 
= \acscA = \E. 

That is, the radius of the nine-points circle of any triangle 
is one-half the radius of the circumcircle of the triangle. 

All other properties of the nine-points circle of O x 2 3 
(circumcircle of ABC) can be reduced to properties of the 
nine-points circle of ABC (circumcircle of P x P 2 Pn) by the 
same substitutions. 

If F is the center of the circumcircle (see Fig. 58), 
K = area BFC 4- area CFA + area ^LFB ; 
that is, rs = 1 all cos A + \ bR cos B + i cE cos C. 

rr 2S 2ttR 

Hence = — 

a cos A+b cos B + c cos C 2 irr 

That is, the perimeter of the triangle is to the perimeter 
of the pedal triangle as the circumference of the circum- 
circle is to the circumference of the incircle. 



104 PROPERTIES OF TRIANGLES. 

50. Abscissae and ordinates of various centers. — Count 
abscissae from A toward B. 

For F, the center of the circumcircle. 

The abscissa is \ c and the ordinate is \ c cot C. 

For 0, the center of the incircle. 

The abscissa is s — a and the ordinate is (s — a) tan ^ A 

For O x , the center of the excircle in A. 

The abscissa is s and the ordinate is s tan \ A. 

For 0. 2 , the center of the excircle in B. 

The abscissa is — (s— c) and the ordinate is (s — c) cot \ A. 

For 3 , the center of the excircle in C. 

The abscissa is s — b and the ordinate is — (s— b) cot J A. 

For H, the intersection of the altitudes (prthocenter). 

The abscissa is b cos A and the ordinate is b cos A cot B. 

For 6r, £7ie intersection of the medians. 

The abscissa is \ (b cos A+c) and the ordinate is ^ 6 sin A 

For N, the center of the nine-points circle. 

The abscissa is \ b cos A + \ c and the ordinate is \ b sin ^1 
— \ c cot C y . 

To obtain the abscissa and ordinate of N, note that the 
angle BP ?J N is 90° — (A — B). Then these values can be 
deduced from (33) and (34). 

To obtain the distance between any two of these points, sub- 
tract their abscissa and square the difference ; subtract their 
ordinates and square the 
difference. The square 
root of the sum of the 
squares gives the required 
distance. 

These distances can also 
be obtained from the tri- 
angle. Thus to find the 
distance FH, find AH, AF, angle FAH, and apply (20) : 

AH =bcosA sec (90° - B) (See Fig. 61) 

5= b esc B cos A = a esc A cos A = a cot A. 




DISTANCES BETWEEN CENTEKS. 105 

Since a = 2 R sin A, AH = 2 R cos A. 
Angle BAH = 90° - B ; BAF = 90° - C; 
hence angle FAH= C — B. 

Hence .F.ff 2 = R 2 [1 + 4 cos 2 .4 - 4 cos (C - 5) cos A]. 

In (34) put 90° - C for A, 90° - B for £, and 180° - A 

for r* 

u °* cos (C - jB) = 2 cos C cos £ + cos A 

Hence FH 2 = R 2 (l-8 cos ^L cos B cos C). 

To find OF and 0/". 
AO=rcsci^L and angle FAO=±A-(90-C)=±(C-B). 
OF 2 =R 2 + r 2 csc 2 ±A - 2 ifr esc i A cos J (C - B). 
Since r = 4 R sin i J. sin i 5 sin i (7, 

r 2 esc 2 1 A = 4 Rr esc J A sin -i .B sin | C = 2 ^ r 2 ( s ~ a ) . 

By (32), esc i J. cos \ (C - B) = ^±£ 

Substituting O.F 2 = R 2 - 2 Rr. 

Similarly, O x F 2 = R 2 + 2 B^. 

Exercises. — Show that Oi7 = 2 r 2 —±R 2 cos J. cos jB cos C. 
ON=\R-r. 
1 N=±R + r 1 . 

PROBLEMS. 

51. 112. The radius of the incircle of a triangle is 6; 
B = 38° 53', C = 73° 44'. A line is drawn across angle A 
parallel to the side BC and tangent to the incircle. In the 
small triangle thus formed, a circle is inscribed. Find the 
radius of this latter circle. 

113. In the triangle of the last problem (r=6, A=67° 23% 
a small circle is drawn touching each side of angle A, and 



: 



PROPERTIES OF TRIANGLES. 



also tangent externally to the incircle. Find the radins of 
this small circle. 

Ul In the same triangle (r = 6, A = 67° 23' f B = 38 3 o3% 
the incircle touches the sides at A^ 2? r and C v Find the 
radins of the incircle of the triangle A^ B^ C v 

Maliatti's problem. — T : :::-:: . - ::: :. nir^i'Ir ::.itt :■:: :"---. 
each touching the other two and also two sides of the tri- 

Let the sides of the triangle 
be a = 14S, 6 = 175. ' =153, 
and the angles A = 53° 8', 

= ~1 4', = 48'. 

Denote the radii of the Mal- 
fatti circles bv x. »/. and z. 



EF'=DE--1jF- 

= x+yf- ^-/0"=^a^ 

Hence 
.rcotA4-|-2V^-r-yeot^2>=c. 

Multiply by sin \ B cos ^ B = I sin 5. 
/ ::: r /f sin t 5ccs r 5-2sinr£cos r ^ Jr/-/cos 




B = ^csinB. 



* Sterner has given the following 
construction for this problem. Let 
O he the center of the incircle of 
ABC. Through the point of con- 
tact D of the incircle of AOB, draw 
a tangent to the incircle of BOG, 
iitt:— _ L _" ::. F --- i- :ir:lr : 
the triangle 5i>F is a 3Ialfatti cir- 
cle. Each of the other two Malfatti 
circles is constructed in a similar 




MALFATTI'S PROBLEM. 107 



The coefficient of x equals sin 2 \B -\ ; 

(X 

also c sin B = p v Hence 

(Vx sin \ B + Vy cos J £) 2 = ^p 1 - ( s ~ a ) x . 

When y is changed to z, B is changed to C. 

(V^sin| + Vzcosi C) 2 = ipj - (s ~ a)a? . 

Equating and changing letters, 

V» sin | B + Vy cos ^ B = V# sin J C + Vz cos J C. 

Vy sin ^ C + Vz cos ^ C = Vy sin J ^4 + V» cos J A 

Adding these and arranging, 

Vy _ cos i A — sin \ B + sin \ C _ qqfi 
■y^ — cos £ 5 — sin ^ A + sin £ C ~~ 

Substituting, a = 30, y = 26.28, 2 = 28.61. 

Properties of the Malfatti radii. — In the equation* 

sin A + sin B + sin C = 4 cos |- A cos £ -B cos % C, 

put 90° — | ^4 instead of A, —\B instead of B, and 180° 
— % C instead of C ; then change A and B : 

cos \A — sin \ B + sin ^- O = 4 cos (45° — \ A) cos £ I? sin £ (7, 

cos ^ 5 — sin J J. + sin | C = 4 cos -£■ A cos (45° — J B) sin J- (7. 

Hence Vy = cos ( 45° - j .4 ) cos j i? 

V^ cos 1^4 cos (45° -\B) 

Now cos (45° — \A) = cos 45° (cos i A + sin i J.),| 

Also cos (45° -|5) = cos 45° (cos \ B -f sin J 5). 

Hence 

Va?(l + tan 1^4) = Vy (1 + tanJ.B) = Vz(l + tan \ G). 

* Proved on next page. 

t See equation (68), Appendix II, page 145. 



108 PROPERTIES OF TRIANGLES. 

52. Relations between the angles of a triangle. — When two 
relations connecting the same two segments are obtained, 
their comparison gives a relation between the angles of the 
triangle. 

For example, E (sin A + sin B + sin C) = s. 

s = 4c E cos ^ A cos \ B cos ± C. 
sin A + sin B -f sin C = 4 cos ^ J. cos i JB cos i (7. 
And since the only condition on A, B, and C is that their 
sum is 180°, this equation is true for any three angles having 
this sum. Thus, apply it to the pedal triangle, 

180° -2 A for A, 180° - 2 B for B, 180° -20 for C. 

sin 2 JL + sin 2 5 -f sin 2 (7 = 4 sin A sin 5 sin (7. 
By adding and multiplying equations in (22), 

r r 

ILL y3 rfi ^ 

Hence 

cot \ A 4- cot \ B + cot i (7 = cot i ^4 cot 1 5 cot i (7. 
In this, put 
180° - 2 .4 for Ay 480° -2B for B, 480° - 2 (7 for (7. 

tan A + tan jB -f- tan C = tan J. tan B tan (7. 
Divide this equation by the right-hand member, 

cot B cot C + cot C cot ^4 + cot A cot 5 = 4. 

It will be seen that there is no limit to the number of re- 
lations which can thus be derived. 

Exercises. — For all triangles, prove 

sin 2 A + sin 2 B + sin 2 C = 2 + 2 cos A cos B cos C, 
sin 2 A + sin 2 B + cos 2 0=1 + 2 sin A sin 5 cos C, 
cos ^4 + cos B + cos (7 = 4 4- 4 sin i A sin 1 5 sin i (7, 
cos 2 J. + cos 2 5 + cos 2 (7 = — 4 — 4 cos J. cos B cos (7. 



ELEMENTS OF OBLIQUE TRIANGLES. 



109 



Exercises. — The following table contains elements of six 
different triangles. Assume three independent elements in 
any one of the triangles and compute other elements. 





l 


2 


3 


4 


5 


6 


a 


13 


25 


37 


37 


68 


68 


b 


15 


26 


13 


15 


65 


75 


G 


14 


17 


40 


44 


57 


77 


<h 


9 


10 


5 


9 


25 


45 


c 2 


5 


7 


35 


35 


32 


32 


s 


21 


34 


45 


48 


95 


110 


s—a 


8 


9 


8 


11 


27 


42 


s-b 


6 


8 


32 


33 


30 


35 


s—c 


7 


17 


5 


4 


38 


33 


A 


53° 8' 


67° 23' 


67° 23' 


53° 8' 


67° 23' 


53° 8' 


B 


67° 23' 


73° 44' 


18° 55' 


18° 55' 


61° 56' 


61° 56' 


C 


59° 29' 


38° 53' 


93° 42' 


107° 57' 


50° 41' 


64° 56' 


K 


84.00 


204.0 


240.0 


264.0 


1710 


2310 


R 


8.125 


13.54 


20.04 


23.12 


36.83 


42.50 


r 


4.000 


6.000 


5.333 


5.500 


18.00 


21.00 


n 


10.50 


22.67 


30.00 


24.00 


63.33 


55.00 


r 2 


14.00 


25.50 


7.500 


8.000 


57.00 


66.00 


r 3 


12.00 


12.00 


48.00 


66.00 


45.00 


70.00 


Pi 


12.92 


16.32 


12.97 


14.27 


50.29 


67.94 


P>2 


11.20 


15.69 


36.92 


35.20 


52.62 


61.60 


Ps 


12.00 


24.00 


12.00 


12.00 


60.00 


60.00 


(h 


12.95 


17.11 


16.33 


20.01 


50.53 


67.96 


<h 


11.22 


16.19 


37.92 


39.65 


53.18 


61.93 


% 


12.09 


24.04 


13.16 


12.56 


60.07 


60.18 


m 1 


12.97 


22.49 


23.29 


27.17 


50.80 


68.09 


m 2 


11.24 


20.33 


37.98 


39.95 


53.67 


64.02 


m 3 


12.17 


24.19 


19.21 


17.69 


60.10 


61.39 



CHAPTER VII. 

SPHERICAL TRIANGLES. 

53. If there are three points on the surface of a sphere, 

and each pair of points is joined by an arc of a great circle, 

the surface of the sphere is divided into two parts, each of 

which is called a Spheri- c 

cal Triangle. Join each ^^^\\ 

of the three points to 0, ^^^ \ N* 

the center of the sphere. ^^^^ j \ 

The arcs joining the CK~-- -)b 

pairs of points on the ^^\^^ I / 

surface measure respec- ^^^^^^ / / 

tively the plane angles ^"""\K 

formed at the center of 

Fig. 64. 
the sphere. Thus the 

arc AB measures the plane angle A OB. Also the angles 

between the arcs are equal respectively to the dihedral 

angles made by the planes of the arcs. Thus the angle A 

of the spherical triangle is equal to the angle between the 

planes A CO and ABO. 

There is no limit to the number of different possible 

values of any side or angle of a spherical triangle. For the 

arc joining A and B may be any arc AB + any number of 

circumferences. And the angle A may be generated by 

turning AG away from AB through any angle A + any 

number of revolutions. Since any function of n x 360° + x 

is equal to that function of x, it follows that any single 

solution of a spherical triangle includes the solution of all 

spherical triangles which have any side or angle increased 

110 



THE GENERAL TRIANGLE. 



Ill 



by any number of times 360°. Therefore, without loss of 
generality, each side and each angle may be considered less 
than 360°. 

Taking each side less than 360°, there may be : 





Fig. G5. 

a = a Aq = 300° - A 
b = b B = 360° - B 
c = c C = 3(30° - C 

1. None more than 180°. 



Fig. 66. 
a x = 360° - a A 1 = 360° - A 
b 1 = b Bi= 180° - B 

Cl = c d = 180° - C 

2. One more than 180°. 




Fig. 67.* 
a 2 = 360°-a A 2 =lS0°-A 
b 2 = 360° - b B 2 = 180° - B 
c 2 = c C 2 = 300° - C 

3. Two more than 180°. 



Fig. 68.* 
a s = 300° - a A s = A 
b s = 300° - b B S = B 
c 3 = 360° - c C 3 = C 

4. Three more than 180°. 



* Out of perspective. It will be well if all the diagrams in this 
chapter are drawn with colored chalk on a spherical blackboard. 



112 



SPHERICAL TRIANGLES. 



The four triangles in 
Eig. 69 are called 
Associated Triangles. 
The three triangles as- 
sociated with ABC are 
obtained by prolonging 
the sides of ABC. 

Spherical triangles, 
like plane triangles, are 
right, isosceles, equi- 
lateral, and oblique ; but 
a right spherical triangle 
may have more than one 
right angle. If one or more sides of the triangle is a 
quadrant, the triangle is said to be quadrantal. 




FORMULAE EOR RIGHT TRIANGLES. 

54. Let ABC be a spherical right triangle, right angled 

at B. Let the angles A and C be acute, and each side of 

the triangle be less than 

180°. Join A, B, and C 

to 0, the center of the 

sphere. Then the planes 

AOB and BOC are at 

right angles. From any 

point D in the edge OC, 

drop a perpendicular A, 

DB l to the plane AOB. A 

Fig. 70 
Through DB l pass a 

plane at right angles to the edge OA. Since DB L is per- 
pendicular to the plane AOB, it is perpendicular to OB x 
and B X A X ; and since A l B l and A X D are both perpendicular 
to OA, the angle B X A X D measures the angle between the 
planes AOB and AOC 




FORMULAE FOR RIGHT TRIANGLES. H3 

OB sin a — B X B = A X B sin A ; and A X B = OB sin b. 
Therefore sin a = sin b sin A. (35) 

OB x tan a = ^D = A X B X tan J. ; and A X B X = OB x sin c. 
Therefore tan a == sin c tan vf. (36) 

OA Y tan 5 = ^Z) = A X B X sec J. ; and A X B X = OA x tan c. 
Therefore tan b = tan c sec A- (37) 

If the angle A 3 B 3 C S , Fig. 69, is a right angle, these for- 
mulae (35), (36), and (37) may be shown to apply to the tri- 
angle A 3 B 3 C 3 . Thus put for a the value 180° — a 3 , put for 
b the value 180° -b 3 , put for A the value 180° -A 3 . There 
is obtained: 

sin (180° - Ob) - sin (180° - b 3 ) sin (180° - A 3 ) ; 
whence sin a 3 = sin b 3 sin A 3 . 

tan (180° - a 3 ) = sin c tan (180° - A 3 ) ; 
whence tan a 3 = sin c 3 tan ^4 3 . 

tan (180° - 6 3 ) = tan c sec (180° - A 3 ) ; 
whence tan b 3 — tan c 3 sec ^1 3 . 

It therefore follows that the formulae (35), (36), (37) are 
true when one of the angles of the right spherical triangle 
is obtuse. In a similar manner, these formulae may be 
shown to apply to the triangle A 2 B 2 C 2 , Fig. 69, and are 
therefore true when two of the angles are obtuse. 

It therefore follows that formulae (35), (36), (37) are true 
for all right triangles in which each of the parts is less 
than 180°. 

In the same three formulae put successively the values 
of the parts of the various triangles in Figs. 66 to 6S, in 
which assume B = 90°. It will be found that all three 
formulae apply ; they are therefore true for all right spheri- 
cal triangles. 



114 



SPHERICAL TRIANGLES. 



Collecting, sin a = sin b sin A, (35) 

tan a = sin c tan A, (36) 

tan b = tan c sec A. (37) 

Changing letters, sin c = sin 6 sin C, (38) 

tan c = sin a tan C, (39) 

tan b = tan a sec C. (40) 

Sq. root product (35) to (40), sec b = tan A tan C. (41) 

(41) -h product (36) and (39), sec b = sec a sec c. (42) 

(35) x (40) -- (42) gives sec c = sin A sec C. (43) 

(37) x (38) -T- (42) gives sec a = sec A sin 0. (44) 

55. The following diagrams with the accompanying rules 
will be found useful in rapidly picking out the relation 
which exists between any three parts of a spherical right 
triangle. 



— sin A- ~$ec Qr-tan 




Fig. 71. 



Eule I. Any function 
is equal to the product of 
its adjacent functions. 

Rule II. Any function 
is equal to the product of 
adjacent function and the 
reciprocal of the next func- 
tion. 




sec A 



Fig. 73. 



SOLUTIONS. 115 

56. In the discussion of plane right triangles it was found 
that any two parts, provided one of them was a side, was 
sufficient to make the solution of a right triangle deter- 
minate. In spherical triangles with one right angle, there 
will always be two solutions (if any) for given values of 
two given parts ; and therefore in addition to values of two 
parts there must be included in the data a condition by which 
the quadrant of any one of the three unknowns may be deter- 
mined. As each unknown is determined by a function, the 
algebraic sign of any other function (except the reciprocal 
function) will be sufficient to determine the quadrant of 
such an unknown. 

Given a = 86° 40', c = 32°40', A acute ; find A and b. 

tan A = tan a esc c sec b = sec a sec c 

1.2348 1.2355 

0.2678 0.0748 

1.5026 1.3103 

.4 = 88° 12' b = 87° 12' 

The value of b cannot be 272° 48 ', since the fourth quad- 
rant is inadmissible by either of the two formulae : 

sin b = sin a esc A and tan b = tan c sec A. 

Had A been known to be greater than 90°, the required 
values would have been A = 268° 12' and b = 272° 48'. 

Given A = 213°, b = 54° 20', and C known to be greater 
than 180° ; find all the parts. 

tan C = sec b cot A sin a = sin b sin A tan c = tan b cos A 
0.2343 9.9098 0.1441 

0.1875 9.7361 n 9.9236 n 

0.4218 9.6459 n 0.0677 n 

C = 259° 16' a = 333° 44' c = 229° 27 ' 

Quadrant given by sec a = sec A sin C, sec c = sin A sec C; 
or tan a = tan b cos C, sin c = sin b sin C. 



116 



SPHERICAL TRIANGLES. 



Given b = 53° 12', C= 110° 48', c acute ; find a and A. 



tan a = tan b cos (7 

0.1260 
9.5501 n 



9.6764 n 
a = 334° 37' 



tan A= secB cot C 

0.2226 
9.5796 n 
9.8022 n 
.4 = 327° 37' 



Given .4 = 327° 37', C = 110° 48', 6 more than 180°. 
sec b = tan A tan C sec a = sec ^4 sin sec c = sin A sec (7 

9.8022 n 0.0734 9.7288 ra 

0.4204 71 9.9707 0.4496 n 

0.2226 = 0.0441 + 0.1784 

b =306° 48' a = 25° 23' c = 3H°32' 

Given J. = 42° 20', C=353°50'. Impossible. Why? 
Given A = 42° 20', a = 268°. Impossible. Why? 

Given 6 = 98°, a = 268°. Impossible. Why? 

Assume the values of any two parts of any triangle in 
any row of the following table and compute the value of 
any other part in that row : 



A 


C 


a 


6 


c 


61° 35' 


30° 29' 


20° 17' 


23° 13' 


11° 32' 


61° 35' 


30° 29' 


339° 43' 


336° 47' 


348° 28' 


118° 25' 


329° 31' 


339° 43' 


336° 47' 


11° 32' 


298° 25' 


149° 31' 


339° 43' 


23° 13' 


11° 32' 


63° 15' 


135° 34' 


50° 0' 


120° 56' 


143° 5' 


296° 45' 


64° 26' 


310° 0' 


120° 56' 


143° 5' 


116° 45' 


224° 26' 


310° 0' 


239° 4' 


143° 5' 


63° 15' 


135° 34' 


310° 0' 


239° 4' 


216° 55' 



POLAR AND QUADRANTAL TRIANGLES. 



117 



57. Polar triangles. — Let ABC be a spherical triangle 
with each part less than 180°. With A, B, and C, respec- 




tively, as poles, describe great circles B'C, C'A', and A'B'. 
Now if with A', B', and C as poles, great circles are de- 
scribed, triangle ABC is obtained. The two triangles ABC 
and A'B'C, thus related, are said to be polar to each other. 
It is proved in geometry that the relations of their parts are : 

a =180° -A'; b = 180°-B'; c = 180 o -C". 

^ = 180°-a'; B = 180°-b'; C=180°-c'. 

Quadrantal triangles. — If the triangle ABC is a right 
triangle, its polar triangle A'B'C is a quadrantal triangle. 
Hence, substituting in (35), (36), and (37) : 

sin (180° - A') = sin (180° -B') sin (180° - a 1 ) ; 
or sin A' = sin B' sin a'. (45) 

Similarly, tan A' = sin C tan a' ; (46) 

tan B' = — tan C sec a'. (47) 

These formulae are true for all quadrantal triangles hav- 
ing each part less than 180°. 



118 SPHERICAL TRIANGLES. 

Now assume b, Fig. 66, to be a quadrant. 

In (45), (46), and (47), put 360° - A x for A', 180° - B, for 
B', 180° - d for C", and 360° - a x for a'. After reduction 
there is obtained : sin A x = sin B x sin a x ; tan J. x = sin C^ tan a ; 
tan B x = — tan Ci sec a x . 

Hence the formulae (45), (46), (47), are true for a quadran- 
tal triangle with one angle greater than 180°. 

The triangle made up of the hemisphere + the triangle 
ABC, Fig. 66, has its angles A, 180° + B, and 180° + C, and 
its sides b, c, and 360° — a. These values also satisfy for- 
mula* (45), (46), (47). The values of the parts of the 
triangle A B C , Fig. 65, also satisfy the same formulae. 
Hence these formulae are true for all quadrantal triangles. 
The formulae for the solution of a quadrantal triangle are : 
(b = 90°) : 

sin A = sin B sin a, sin C = sin B sin c, 

tan A — sin C tan a, tan G = sin A tan c, 

sec A = sin c sec a, sec C = sin a sec c, 

tan B = — tan ^L sec c, sec B- — tan a tan c, 

tan B = — tan (7 sec a, sec B = — sec A sec O. 

They are obtainable from the formula* for right triangles by 
interchanging the large and small letters and prefixing a minus 
sign to every function except the sine and cosecant. 

Isosceles triangles are solved by drawing a perpendicular 
from the unequal angle upon its opposite side, thus forming 
two right triangles. 

Regular polygons are solved by joining the center of the 
polygon to each vertex and also to the middle of each side, 
thus forming 2 n equal right triangles, n being the number 
of sides. 



ELEMENTS OE SPHEKICAL FIGURES. 



119 



Elements of Some Quadrantal Triangles. 



a 


b 


c 


A 


B 


C 


240° 0' 


90° 


45° 0' 


225° 0' 


125° 16' 


144° 44' 


103° 26' 


90° 


154° 0' 


238° 0' 


240° 41' 


202° 28' 


292° 22' 


90° 


48° 50' 


300° 22' 


68° 55' 


135° 23' 


104° 53' 


90° 


133° 40' 


249° 12' 


255° 18' 


224° 24' 


174° 13' 


90° 


94° 8' 


175° 57' 


135° 34' 


135° 43' 


104° 53' 


90° 


133° 40' 


110° 48' 


104° 42' 


135° 36' 


299° 47' 


90° 


124° 51' 


307° 15' 


113° 29' 


48° 48' 



Elements of Some Isosceles Triangles. 



a 


b 


c 


A 


B 


C 


54° 20' 


54° 20' 


87° 5' 


47° 0' 


47° 0' 


116° 0' 


63° 56' 


63° 56' 


302° 0' 


105° 44' 


105° 44' 


294° 40' 


112° 48' 


112° 48' 


254° 9' 


56° 12' 


56° 12' 


240° 8' 


100° 4' 


100° 4' 


207° 2' 


42° 24' 


42° 24' 


198° 8' 


127° 50' 


127° 50' 


272° 52' 


42° 24' 


42° 24' 


238° 30' 


101° 7' 


101° V 


303° 31' 


83° 57' 


83° 57' 


302° 20' 



Elements of Some Regular Polygons. 



n 


a 


G 


n 


a 


C 


3 


60° 


70° 32' 


9 


20° 


145° 11' 


4 


90° 


90° 0' 


10 


30° 


159° 53' 


4 


60° 


109° 28' 


12 


24° 


161° 52' 


3 


108° 


116° 34' 


15 


20° 


166° 40' 


5 


60° 


138° 11' 


18 


18° 


171° 14' 


6 


50° 


145° 42' 


20 


12° 


166° 34' 


8 


40° 


158° 57' 


30 


10° 


173° 22' 



120 



SPHERICAL TRIANGLES. 



OBLIQUE SPHERICAL TRIANGLES. 

58. As in plane triangles, the solution is effected by- 
dropping a perpendicular from a vertex upon the opposite 
side. Two right triangles are thus formed, in each of which 
a value of the perpendicular is found. The solution is then 
obtained by equating the values of the perpendicular. 

This method fails when the given parts are the three 
sides or the three angles. 

The solutions for these two cases will now be found. 




Fig. 75. 



59. The in- and excircles. — Prolong the sides AB and AC, 
and bisect the angles A, B, and CBC X by great circles. 
From the intersections and 1 drop perpendiculars on AB. 
With and O x as poles, describe small circles with radii 
OC and OC ± respectively. It is easy to see that the various 
triangles AOB and AOC , AO^ and AO^, etc., are equal 
in pairs ; therefore the small circles touch AC in B and 
B x , and BC in Aq and At. 

Then exactly as in plane triangles it may be shown that 
AC X = \ (a -f b + c) = s, AC = s - a, BC = s- b, BC\ 
= s — c, etc. 

Since the angles ABC + CBC, = 180°, the sum of their 
halves, or OBO x = 90° ; therefore, O.BC = 90° - i B. 



THE IN-, EX-, AND CIRCUMCIRCLES. 



121 



In triangle AO x C^ tan \A — esc s tan r v 

In triangle AOC Q , tan r = sin (s — a) tan -J- A (48) 

In triangle BOC , tan r = sin (s — b) tan i jB. 

In triangle BOyC^, tan r x = sin (s — c) cot i _B. 

Hence tan r = Vcsc s sin (s — a) sin (s — 6) sin (s — c). (49) 

60. The incircle of the polar triangle. — Since (48) is true 
for all triangles, it is true for the polar of ABC. That is, 
tan r' = sin (s' — a') tan \ A!. 

Now A' = 180° - a and \ A' = 90° -\a. 

2 s' = 180° - ^ + 180° - £ + 180° - C ; hence s' = 270° - S, 
where 6" stands for %(A + B+C); since a' =180°— -4, 
s' - a' = 90° - (£ - A). Substituting these values in (48) 
and taking the reciprocal of the resulting equation, 

cot ?*' = tan i a sec (S — A). (50) 

The same process applied to (49) gives 

cot r' = V- cos S sec (S - A) sec (S - 2*) sec (£ - C). 

The circumcircle. — Bisect 
the sides of the triangle 
ABC by perpendiculars 
meeting in F. On account 
of the equality of the tri- 
angles in pairs, FA = FB 
= FC, which denote by R. A 
Also the angles of the isos- 
celes triangles are equal in 
pairs. Denote FBC and 
FCB by x, FCA and FAC 
by y, F^B and FBA by z. 

Since x + y = C, y -\-z — 

In triangle BM X F, tan R = tan \ a sec (# — -4). (51) 




By (50), 



tan R = cot r'. Hence 



tan R = V - cos £ sec (£ - A) sec (£ - 5) sec (S - C). (52) 



122 



SPHERICAL TRIANGLES. 



61. Equations (35) to (44) applied to Fig. 77 give 
tan Cj = tan b cos A and tan c 2 = tan a cos B. 
tan d = sec b cot A and tan C 2 — sec a cot 5. 
sin p = sin b sin ^4 = sin a sin jB. 
sin p = tan c x cot C x = tan c 2 cot C 2 . 
tan p = sin c x tan ^4 = sin c 2 tan 5. 
tan p — tan 6 cos C\ = tan a cos 2 . 
sec p = sec b cos c x = sec a cos c 2 . 
sec p = sec A sin Ci = sec B sin 2 . 
Eearranging (55) and changing letters, 

sin a esc A = sin b esc J5 = sin c esc O. 




(53) 
(54) 
(55) 
(56) 
(57) 
(53) 
(59) 
(60) 

(61) 



62. Methods of solving oblique spherical triangles. — The 

following methods may be modified in various ways espe- 
cially to suit problems in which all of the unknown parts 
are not desired. 

(1) Given two sides and the angle opposite one of them, say 
a, b, A. Find B by (55), c by (53), and G by (54) or (61). 



SOLUTIONS. 123 

(2) Given two angles and the side opposite one of them, say 

A, B, b. Find a by (55), c by (53), and C by (54) or (61). 

(3) Given two sides and the included angle, say b, c, A. 
First find c x by (53) ; B can be found by (57), and a by (59). 
If C is desired, it can be found by (61). When only one 
of the unknown angles is desired, draw the perpendicular 
through the angle which is not desired. 

(4) Given two angles and the included side, say b, A, O. 
First find C x by (54) ; a can be found by (58), B by (60), 
c by (61). If only one of the unknown sides is desired, 
draw the perpendicular upon the side which is not desired. 

(5) Given the three sides. — The angles can be found by 

(49), (48), etc. 

(6) Given the three angles. — The sides can be found by 
(52), (51), etc. 

SOLUTIONS. 

63. 1. Given two sides and the angle opposite one of them, 

a = 236°5'2, & = 93°40', .4 = 220°, and B known to be be- 
tween 90° and 180° : find c and B. 



tan c x = tan b cos A 


cos c 2 = sec b 


cos Cj cos a 


log tan & 1.1933 n 


log sec b 


1.1941 n 


LeosA 9.8843 n 


L cos c x 


8.9211 


log tan c : 1.0776 


L cos a 


9.7377 n 


h == 85° 13', c 2 = 315°27'. 


L cos c 2 


9.8529 


sin B = sin A esc a sin b 


Check : tan c 2 


= tan a cos 


LsinA 9.8081 n 


L tan c 2 


9.9932 n 


log esc a 0.0771 n 


log tan a 


0.1853 


L sin b 9.9991 


LcosB 


9.8081 n 


Zsin£ 9.8843 


c = 40° 40', 


B = 130°. 



1-24 SPHERICAL TRIANGLES. 

2. Given two angles and the side opposite one of them, 
A = l 30°, B=110° 3 b = 126°58'; find C (less than 90°). 

tan C\ = sec b cot A sin C> = see A sin C\ eos 5 
I jg se : 2209 n log sec ^ 0.1919 

L ::~A AAA l si — : 

log tan C t 0.1447 LcosB jU»341 

n _ 54« _ _ Z sin C 2 9.6360 

C. = 25° 38', and hence (7 = 80°. 

3. Given two sides and the included angle, :< = 2ol : ov 
e=275° 1 . ^=132° 30', and a less than ISO 3 ; find a and 5. 

tan c a = tan 6 bos A\ :■-_ = : — o- ; 

sec a = sec & cos ; e : :_ \ tan j3 = ran ^1 sin c, esc Cj. 

log tan 5 0.4755 log sec 6 0.4985 n log tan A 0.0379 

Zcos-I 9.8297 n Zcosc, 9.6470 n isinq 9.9524 

log tan q 0.3052 n log sec c, 0.0303 n log esc c, 0.4424 

. = 116 . _ log sec I""- n logtsmB ".A A 

c = 158°50' a=131°50' 5 = 290°16 f 



4. Given two angles and the included side, & = 251°30 f , 
^=132° 3" . C7=279° : J . and a less than 180°; find a and ^. 

tan d = sec b cot J. : C-_ = C — O : 

tan a = tan 6 cos (7, sec C 2 ; sec B = sec J. sin d esc C,. 

log sec 6 0.4985 n log tan & 0.4755 log sec ^10.17 

L::'A 9.9'^21 \ Zo:=C : '.:"'. ol4S •■■ X si:: C : ;'.vT.;4". 

g ran C, 0.4606 log sec C 2 O.Oo".. 1 log esc C s 0.3153 

_ : log tan a jM^g2 « log sec # 0.4610 

= 28° 56 a = 131° 50' £=— : 1/ 



SOLUTIONS. 



125 



5. Given the three sides, a = 124°, b = 54°, c = 98°, and 
A less than 180° ; find the angles. 

tan r = Vcsc s sin (s — a) sin (s — b) sin (s — c) ; 



cot! A 



S m( S -q) ;co sin( S -6) - ete 

tanr tanr 



6. Given the three angles, .4 = 220°, 5 = 130°, C = 150°, 
and a known to be more than 180° ; find the sides. 



tan R = V — < 


3os aS sec (S — 


■ A) sec (S - B) sec (S - C) ; 


cot i a = 


_sec(S-A)^ 
tan R ' 


etc. Check by (61). 


Computation for Problem 5. 


Computation for Problem 6. 


log esc 138° 


0.1745 


L{- cos 250°) 9.5341 


L sin 14° 


9.3837 


log sec 30° 0.0625 


L sin 84° 


9.9976 


log sec 120° 0.3010 n 


L sin 40° 


9.8081 


log sec 100° 0.7603 n 


L tan 2 r 


9.3639 


log tan 2 R 0.6579 


L tan r 


9.6819 


log tan R 0.3289n 


L cot \ A 


9.7018 


Xcot^a 9.7336n 


log cot \ B 


0.2957 


ZcoHfr 9.9721 


log cot J C 


0.1062 


log cot Jc 0.4314 


J. = 126° 34', 


B = 53° 42', 


a = 236° 52', b = 93° 40', 


and = 


: 76° 6'. 


and c = 40° 38'. 



If A is known to be greater 
than 180°, the negative root 
of tan 2 r would have been 
taken. Hence we would find 

A = 233° 26', B = 306° 18', 
C = 283° 54'. (See Fig. 65.) 

As a check, apply (61). 



L sin a 
log esc A 
isin b 
log esc 5 
L sin c 
log esc C 

Check 



9.9229 n 

0.1919 n 

9.9991 

0.1157 

9.8137 

0.3010 

0.1148 



126 SPHERICAL TRIANGLES. 

64. It is frequently convenient to express a desired part 
directly in terms of given parts, especially when a chain of 
triangles is to be solved. It is recommended that the fol- 
lowing formulae be derived from a diagram, in the manner 
that the formulae for right triangles were derived. 

They will here be obtained by applying equations (67) 
and (68) proved in Appendix II., page 145. 

Thus, from (59), (68), and (53), 

cos a — cos b sec c x cos (c — c : ) 

= cos b sec c 2 (cos c cos q + sin c sin c x ) 
= cos b cos c + cos b sin c tan c x 
= cos b cos c + cos b sin c tan b cos A ; 
cos a = cos b cos c -1- sin b sin c cos A. 
Multiply by esc b esc c, and transpose : 

cos A = — cot 6 cot c + esc b esc c cos a. 
Apply the last two equations to the polar triangle : 
cos A = — cos B cos C 4- sin B sin C cos a, 
cosa= cot B cot (7 + esc B esc C cos A. 
Using in the same manner (57), (67), and (53), 
cot B — — cos c cot A + sin c esc ^4 cot b, 
cot 6 = cot c cos ^1 + esc c sin ^1 cot i?. 

By changing letters, similar relations between other parts 
of the triangle may be obtained. 

The application of these forms involves the use of both 
tables, — the table of logarithms of functions in computing 
the values of each term of the right-hand member in any 
equation and the table of natural values of functions in 
finding the answer. 

As an illustration of the application of these formulae, 
find a (less than 180°) given 

b = 127°, c = 57°, A = 130°. 



ELEMENTS OF OBLIQUE TRIANGLES. 



127 



cos a = cos b cos c + sin b sin c cos A. 

Laosb 9.7795 n 
L cos c 9.736.1 



cos b cos c = - .3278 

sin b sin c cos .4 = — .4305 X sin b 

nat. cos a = — .7583 L sin c 

Therefore a = 139° 19' icos^l 



9.5156 h 
9.9023 
9.9236 
9.8081 n 
9.6340 7i 



With the same data, find B (also less than 180°). 
cot B = — cos c cot A -f- sin c esc ^4 cot 6. 



.4570 

.8249 



- .3679 

B = 110° 12' 



L ( — cos c) 

iyCOtJ. 

i sine 
log esc ^1 
Zcot6 



9.736l7i 
9.9238 71 
9.6599 
9.9236 
0.1157 
9.8771 n 
9.9164 71 



Elements of Some Oblique Spherical Triangles. 



a 


b 


c 


A 


B 


C 


131° 35' 


108° 30' 


84° 47' 


132° 14' 


110° 11' 


99° 42' 


124° 13' 


125° 42' 


82° 48' 


127° 22' 


128° 42' 


107° 33' 


134° 16' 


150° 57' 


55° 42' 


120° 48' 


144° 23' 


97° 43' 


126° 25' 


138° 32' 


45° 54' 


261° 16' 


234° 25' 


298° 6' 


309° 48' 


116° 45' 


129° 12' 


59° 4' 


274° 23' 


300° 5' 


120° 56' 


59° 4' 


106° 10' 


116° 45' 


63° 15' 


91° 7' 


113° 4' 


111° 41' 


79° 35' 


110° 50' 


109° 17' 


87° 35' 


126° 


152° 0' 


75° 0' 


142° 24' 


159° 16' 


133° 14' 


90° 53' 


117° 49' 


132° 5' 


120° 0' 


130° 0' 


140° 0' 


155° 5' 


147° 6' 


33° 2' 


133° 18' 


110° 10' 


70° 21' 



128 SPHEEICAL TRIANGLES. 

THE CELESTIAL SPHERE. 

65. The concave surface on which the heavenly bodies 
seem to be situated is called the celestial sphere. 

Intersections of the celestial sphere: 

I. Its intersection with the plane of the earth's equator 
is called the celestial equator. 

II. Its intersections with the prolongations of the earth's 
axis are called the poles of the heavens. 

III. Its intersection with a plane tangent to the earth 
through the foot of an observer is called the horizon of the 
observer. 

IV. Its intersections with the vertical line through an 
observer are called the zenith and nadir respectively. 

Great circles through the poles are called declination cir- 
cles (or hour circles). On account of the motion of the earth 
on its axis, these circles seem to move from east to west 15° 
every hour. 

Great circles through the zenith and nadir of an observer 
are called vertical circles. The vertical circle through the 
east and west points of the horizon is called the prime verti- 
cal of the observer. The vertical circle through the north 
and south points passes also through the poles ; it is called 
the meridian of the observer. 

The points of intersection of the celestial equator with 
the path of the sun are called equinoxes. The vernal equi- 
nox is passed by the sun on March 21, and the autumnal 
equinox is passed on September 21. 

The arc of the celestial equator from the vernal equinox 
to the declination circle through a heavenly body, say a star, 
is called the right ascension of the star. The right ascen- 
sion is usually expressed in terms of time (15° =1 hour) and 
is measured toward the east. The right ascension of the 



THE CELESTIAL SPHERE. 



129 



meridian at any time is given by a sidereal clock set for that 
meridian. 

The angle which the declination circle makes with the 
meridian (= arc on the equator from meridian to declination 
circle) is called the hour-angle of the star. The arc of the 
declination circle from the celestial equator to the star is 
called the declination of the star. The declination is positive 
Avhen measured from the equator to the north. 

The arc of the horizon from its south point to the vertical 
circle through a star is called the azimuth of the star. The 
arc of the vertical circle from the horizon to the star is 
called the altitude of the star. The azimuth is positive 
when measured in the direction SWNE, and the altitude is 
positive when measured toward the zenith. 

These definitions are illustrated in the following diagram. 










E 

Fig. 78. 


.n. 




N 


north 


QZ 


latitude 


VI) 


right ascension 


P 


pole 


EQW 


equator 


QPD 


hour angle 


Z 


zenith 


SiWNE 


horizon 


DS 


declination 


V 


equinox 


EZVV 


prime vertical 


SiWNA 


azimuth 


s 


star 


NPZSi 


meridian 


AS 


altitude 



The triangle PSZ is sometimes called the Astronomical 
Triangle. Its parts are : 

Arc PZ = arc QP- arc QZ= 90° - latitude. 
Arc ZS = arc AZ — arc AS = 90° - altitude. 
Arc SP = arc DP - arc DS = 90° - declination. 



130 SPHERICAL TRIANGLES. 

Angle PZS = arc S x WNA - arc 8 X WN = azimuth - 180°. 
Angle ZSP is called the parallactic angle of the star. 
Angle ZPS = arc QD = honr-angle. 

Notation. — The hour-angle will be denoted by P, the 
declination by d, the azimuth by Z, the altitude by a, the 
latitude by I 

APPLICATIONS TO GEOGRAPHY. 

115. Find in geographical miles (60 miles = 1°) the 
shortest distance on the surface of the earth from East- 
port (10° E. longitude, 44° 50' K latitude) to Point Barrow 
(80° W. longitude, 71° 20' K latitude). 

116. Find the shortest distance on the surface of the 
earth from Quito (2° W. longitude, 0° latitude) to Point 
Barrow. 

117. A required point is described as being equidistant 
from Eastport, Point Barrow, and Quito. What is the 
distance on the surface of the earth from the point to each 
of the three places ? 

118. Find the distance from Eastport to San Diego 
(40° W., 32° 40' K). 

119. The latitude of Key West is 24° 10' K ; the latitude 
of Attou Island is 53° N. The distance apart of the two 
places is 3722 geographical miles. Find their difference in 
time. 

120. A ship sails from Halifax (lat. 44° 40' K, long. 13° 
25' E.) in a direction N. 81° E., and continues on an arc of a 
great circle for 2472 miles. Find the place arrived at. 

APPLICATIONS TO ASTRONOMY. 

121. Given the latitude of a place and the declination 
of the sun, to find the azimuth of the sun at sunrise and 
the length of the day. 



APPLICATIONS. 131 

Consider the following cases : 
i. I = 40°, d = + 23° 27'; the longest day. 
ii. I = 40°, d — 0° ; days and nights equal. 

iii. I = 40°, d = - 23° 27' ; the shortest day. 

iv. 2 = 66° 33', d = 23°27'; or I and d complementary. 

122. Given the latitude of a place and the declination 
of the sun, to find when the sun is due west (in the prime 
vertical). Consider the following cases : 

i. 1 = 4,0°, d = + 23°27. 
ii. Z = 40°, d = 0°. 
iii. 1 = M° 33', d = + 23°27'. 

123. Given the latitude of a place and the declination 
of the sun, to find the altitude and azimuth of the sun at 
6 o'clock in the morning. Take Z = 40°, cZ = 23°27'. 

124. The interval which elapsed between the transits of 
Eegulus over the prime vertical was observed to be 4 hours 
and 12 minutes. Find the latitude of the place, the decli- 
nation of Eegulus being 12° 40'. 

125. At about 8 o'clock, the altitude of Eegulus was 
observed to be 16° 40'. Three hours and twenty-four min- 
utes later, it was again observed and found to be 40° 20'. 
Find the latitude of the place. 

126. The interval between the observed equal altitudes 
of Antares was 4 hours and 20 minutes. Find the lati- 
tude of the place if the observed altitudes were 54° 20' 
and the declination of Antares is —26° 6'. 

127. From a place north of the equator, the altitude of 
Capella was observed to be 76° 40' when its hour-angle was 
1 hour and 6 minutes, the declination of Capella being 
45° 54'. Find the latitude of the place of observation. 



132 SPHERICAL TRIANGLES. 

A tree 100 feet high stands on a horizontal plane in 40° 
north latitude. What is the area described by its shadow 
between the hours of 9 and 2 on the 21st of March ? 

Let h be the height of the tree, s the length of its 
shadow, Z the azimuth of the sun and of the shadow, P 
the hour-angle, I the latitude, and a the altitude of the sun. 

First, cos Z = — tan a tan I, and by plane trigonometry, 
s tan a = h. The product of these is s cos Z — — li tan I. 
But — s cos Z is the northerly projection of the shadow, and 
h tan I is a constant. Hence, on this day the amount which 
the shadow extends north is a constant. Therefore the path 
of the shadow of the top of the tree is a straight line, due 
east and west, 83.91 feet from the foot of the tree. 

Second, cot Z = sin I cot P, from which the azimuth of the 
shadow at 9 o'clock is 237° 16' and at 2 o'clock it is 138° 4'. 

The area of the triangle passed over by the shadow be- 
tween 9 and 12 is 5477 square feet, and the area described 
between 12 and 2 is 3162 square feet, making the total area 
8639 square feet. 



AREA OF A SPHERICAL TRIANGLE. 

66. Denote the area by K, the half sum of the angles by 
8, the radius of the sphere by R. Then, by geometry, 

#90° 

90° 

Thus, if A = 130°, B = 110°, C = 80° ; K= 2.4435 E 2 . 
Also, if A = 220°, B = 130°, C = 150° ; K= 5.5851 R\ 

128. Given b = 70° 20', c = 38°28', A = 52° 30'.; find K. 

129. Given a = 139°21', & = 126°5S', c = 56°52'; find K. 



APPENDIX I. 



THEORY OF LOGARITHMS. 



67. Arithmetic and geometric series. — Logarithms are a 
series of numbers severally assigned to ordinary numbers, 
and so chosen that the multiplication and division of or- 
dinary numbers may be accomplished by the addition or 
subtraction of their logarithms. 

Thus, if the ordinary numbers are terms in a geometric 
progression (a multiplication series) and to each term we 
assign a term in an arithmetic progression (an addition 
series), it can be shown that the multiplication or division 
of the terms in the geometric progression can be accom- 
plished by the addition or subtraction of terms in the 
arithmetic progression. 



Thus suppose it 
is desired to multi- 
ply the numbers 
r 5 (32) and r 3 (8). 
The logarithm of 
r 5 (32) is a + 5 d (5), 
and the logarithm 
ofr s (8)isa+7d(7). 
Adding these loga- 
rithms gives 2 a 
+ 12 d (12), which 
is larger than the 
logarithm of the 



Numbers. 


Logs. 


r io 


a 


r 9 


a -f d 


r s 


a+ 2d 


r 7 


a + 3d 


r G 


a+ 4d 


r 5 


a -f- 5d 


r 4 


a+ 6d 


r 3 


<x+ Id 


? .2 


a+ Sd 


r 


a+ 9d 


1 


a + 10 d 



Numbers. 


Logs. 


1024 





512 


1 


256 


2 


128 


3 


64 


4 


32 


5 


16 


6 


8 


7 


4 


8 


2 


9 


1 


10 



133 



134 THEOEY OF LOGARITHMS. 

correct product [r 8 (256)] by a + 10 d (10). 
But a + 10 d (10) is the logarithm of 1. 
Heuce log (r 5 x r 3 ) = log ?* 5 + log ?' 3 — log 1. 

log (32 x 8) = log 32 + log 8 - log 1. 

And this relation is perfectly general in any two such 
series. Either by trial or reduction from this relation we 
may obtain 

log (jyq) = logp + log q - log 1, 

log- = log p - log q -f log 1, 

\ogp n = n logp — (?i — 1) log 1, 

log Vp = - (logj? + log 1). 
n 

It will be seen that a logarithm is a function fulfilling 
the condition 

Function (ab) — function (a) -f- function (6) — a constant. 

The theory of logarithms is principally devoted to ob- 
taining the logarithm of any number whatever, — in other 
words, interpolating in the two series. 

Logarithms may be considered geometrically, as they 
were by the inventor of the first system : 

NAPIERIAN LOGARITHMS.* 

68. Suppose two points, a A 

A and _B, to start at the * ' : 

same time with the same 



initial velocity. A starts ° * "• 

at A and moves along the FlG " ,9- 

half-line A A X with a uniform velocity. B starts at B and 

* This name has by some writers been incorrectly applied to 
SpeidelFs Logarithms, which were published in 1619, five years after 
the publication of Napier's logarithmic tables. 



NAPIERIAN LOGARITHMS. 135 

moves along the segment B B Z , with a changing velocity 
such that, when B arrives at any point B x} the velocity of B 
will be proportional to the remaining segment B X B Z . Then, 
if at any time A is at A x and at the same time B is at B x , 
the length of A A X was called by Napier the logarithm of 
B X B Z * 

The numerical value of the length of B B Z Avas taken by 
Napier to be 10 7 , or 10,000,000. It will here be denoted 
by k ; hence from the definition, Napierian log k = zero. 

Suppose that during the first unit of time A moves a dis- 
tance 1, and B moves a distance which denote by (1 — r)1cf 
The distance of the moving point B from B z will then be 
k — (1 — r)k = rk ; and hence Napierian log rk = 1. 

Now during the second unit of time A will move another 
distance equal to 1, making its distance from A equal to 2. 
Since the law of the diminution of the velocity of B remains 
the same throughout its motion, we obtain the distance 
traversed by it in the second unit of time in the manner as 
it was obtained for the first unit of time. That is, multiply 
the distance which it is away from B z at the beginning of 
the unit of time under consideration by the constant 1 — r. 
This gives for the space traveled over by B in the second 
unit of time (1 — r) rk ; so that B is then removed from B z 
by the distance rk — (1 — r) rk = r 2 k. 

Hence Napierian log r 2 k = 2. 

Similarly, Napierian log r % k = 3 ; 

and in general, Napierian log r n k = n. 

* Mirifici Logarithmorum Canonis Constructio. English transla- 
tion by W. R. Macdonald, Edinburgh, 1889. 

t If the initial velocities of A and B are equal as assumed by- 
Napier, and k is very large, the value of r will be nearly equal, but 
slightly less than 1. The distance gone by B in the first unit of time 
will be slightly less than I on account of the retardation of the velocity 
of B as it approaches B z . 



136 THEORY OF LOGARITHMS. 

It will be seen that the numbers form a geometric pro- 
gression and their logarithms form an arithmetic progression, 
and hence that Napierian logarithms are only a particular 
case of the more general logarithms. Xapier solved the 
problem of finding corresponding intermediate terms in 
the two series ; his logarithms are no longer used, but 
the method by which we could interpolate in the Xapierian 
series is quite similar to the method adopted for the same 
process applied to systems now in use. 



PRESENT SYSTEM OF LOGARITHMS. 

69. Suppose two points A and B, moving in the same direc- 
tion, A with a uniform ve- 
locity along the line A A X , 1 ° ** — h 5 -— — 

and B along the half-line . . 

B B X , with a velocity directly £ £ ""^ ST +c *° 

proportional to its distance IG ' " 

from B . If at any time the distance of A from A is A^A^ 
and at the same time the distance of B from B is B B X , 
the length A A X is denned as the logarithm of B B X . 

If it is assumed that A is at zero (A ) when B is at 1 (B } ), 
the logarithm of 1 = 0. This assumption avoids the neces- 
sity of subtracting the logarithm of 1 when multiplying by 
means of logarithms. 

Denote the velocity which B has acquired when at B x by 
v ; then at any other point the velocity of B will be v times 
its distance from B ; thus, the velocity of B at any point 
B x = c times B B X . In the first unit after B leaves B x , A 
will be supposed to move a distance 1, and B will move in 
the same time a distance, which may be denoted by cv times 
B B X \ for a particular system of logarithms, c and v are 
constants. 

Let us count from the instant that A was at A and B at 
B v At this instant B x will coincide with B y , and the dis- 



NATURAL LOGARITHMS. 137 

tance of B from B = 1. In the first unit of time afterward, 
the distance moved over by B becomes cv, and its total dis- 
tance from B at the end of this first unit of time is 1 + cv ; 
therefore, log(l -f cv) = 1. 

Under the same assumptions, in the second unit of time 
A moves another unit. This makes the distance A A X — 2, 
and the motion of B = cv times its distance from B at the 
beginning of this unit of time = cv(l + cv). Therefore, at 
the end of the unit of time, the distance of B from B 

= 1 + cv + cv (1 + cv) = (1 -f- cv) 2 . 

Therefore log (1 + cvf = 2. 

By a similar process of reasoning it may be shown that 
log (1 + cv) 3 = 3, and in general that log (1 + cv) n = n. 

70. It will be seen that the values of the distances of 
A from A at the ends of successive units of time ; namely, 
1, 2, 3 ••• n, form an arithmetic progression: and that the 
values of the distances of B from B ; namely, (1 + cv), 
(1 -\-cv) 2 , (1 + cv) 3 ••• (1 + cv) n , form a geometric progres- 
sion. The value of c depends on the value assumed for v, but 
v is perfectly arbitrary, and values for it may be assumed 
at pleasure ; now a system of logarithms may be constructed 
by assuming for v any value at all, since it has already 
been shown that the terms of any arithmetic progression 
may be used as logarithms of the terms of any geometric 
progression. But in systems of logarithms in actual use, 
special values have been given to v on account of special 
conveniences served thereby. Thus, in higher mathematics, 
the reciprocal of v arises as a factor ; this factor (denoted 
by M) is eliminated and processes simplified by making 
M= 1 (whence also v = 1). The system in which M and 
v are taken as unity is called the Natural System of Loga- 
rithms, and the logarithms in that system are called 
Natural Logarithms, or Speidell's logarithms. 



138 THEORY OF LOGARITHMS. 

71. The number corresponding to a given logarithm. 

Let us now imagine : 

(1) That B, having an initial velocity of v, moves for one 
unit of time divided into n parts. 

(2) That B, having an initial velocity of 1, moves for v 
units of time, each unit being divided into n parts. 

In each case denote the value of the constant, which is c 

for one unit of time, by —for each - of the unit of time. 
m n 

Then under the first assumption the motion of B is 

; and under the second assumption it is f 1 H — 
mj \ m, 

By the binomial formula, 



1+^ 



i + ^y=i+— + ^-i> 2 + ^-i)^-2K +etc> 



mj m 1 • 2 • m 2 1 • 2 • 3 • m 3 

^ . 1 \ nv i . nv , n 2 v 2 — nv . n 3 v s — 3 n 2 v 2 + 2 nv , , 
mj m 1 • 2 • m 2 1 • 2 • 3 • m 3 

Now when n is very great, the distance that B, with a 
velocity v (or 1 under the second assumption) will go from B Y 

in - of a unit of time will approach as a limit - (or - ). But 
n . n\nj 

this distance is also v (or 1) times — . Therefore, when n is 

m 

infinite, m = n. Also, as m and n become very great, 
n(n — l)v 2 -, n 2 v 2 — nv 



and - — both approach 



i ■ 2 m 1 " ' si . I ' ' " rj ! ■':i.m 2 ' 

n 3 v 3 



n(n-l)(n-2)v* d nV-3nV+2nt> ach 

1.2.3-m 3 1.2.3-m 3 FF 1.2.3-m 5 

When these limiting values are put in the values of the 
motions, these values of the motions become identical. 
Denote the common result by b, which is either the motion of 
B from B x for one unit of time with an initial velocity of v, or the 



BASE OF THE NATURAL SYSTEM. 139 

motion of B from Z? x for v units of time with an initial velocity of 1 . 

The value of & is 

6 = 1 + „ +I ^. + _J_ + etc. (62) 

In this value of b, make v = 1 and denote the result by e ; 
e is, therefore, the motion of B from B± while A goes a dis- 
tance equal to 1, both A and B having the same initial 
velocity. The value of e is 

e=1+1+ r i 2 + riT3 +etc - 

The numerical value of e, which is readily computed, is 
2.71828 4- . (See page 10.) 

After B was at B x (or 1) its motion for a unit of time with 
an initial velocity of 1 made its distance from B equal to e; 
hence another unit of time will make its distance 1 x e x e=e 2 ; 
and at the end of v units of time the distance of B from B 
will be e v ; but this = b, the distance that B would be from 
B at the end of one unit of time, if its velocity when at B x 
was v. Therefore at the end of the second unit of time, 
with an initial velocity of v, the distance of B from B 

= 1 x b x b = b 2 = 1 x e v x e v = e 2v ; 

and for n units of time with an initial velocity of v, the dis- 
tance of B from B = b n — e nv ; and hence from the defini- 
tion, b n = e nv is the number whose logarithm is n. That is, 
if .N is a number and n is its logarithm, 

b n = N, where b — e v . 

Taking two numbers, JVi and N 2 , with logarithms n x and n 2 , 

JVi x N 2 = b ni x &" 2 = & ni+ " 2 . 
JSfi-i- N 2 = b" 1 -+- b n * = b ni ~ n *. 

n 

JV> = 6"p and a/^= 6 7 . 



140 THEORY OF LOGARITHMS. 

That is, log (^ x N 2 ) = log N, + log N % 
log (ft -r- JSQ = log N x - log &» 
log N* = p log N 9 
log a/jV = log N-t- r. 
It will be noticed that 
The numbers form a geometric progression ; 
Their respective logarithms form an arithmetic progression; 
These systems of logarithms are only particular cases of 

the more general form of logarithms ; 
Logarithms in these systems satisfy the condition : 
function (ah) = function (a) + function (b) ; 
The constant (log 1) which was introduced in the Napierian 
system will be zero in these systems. 

72. Base. — In the relation b n = N, we see that the loga- 
rithm n is that exponent of a number b corresponding to a 
number N. This is expressed by saying that in these 
systems, a logarithm is an exponent ; the number b which is 
affected by this exponent is called the base of the system 
of logarithms. It is evident that if JV remains constant and 
b is arbitrarily made to change, then n will also change. 

Relations between logarithms in systems with different 
bases. — Let x be the logarithm of a in a system whose base 
is b (written x = log 6 a), and z be the logarithm of b in a 
system whose base is a (written z = log a b). Then b x = a 
and <r = b. 

Now in any of the present systems, taking the logarithms 
of each side of these equations, 

x log b = log a ; z log a — log b ; whence xz = 1. 

That is, log 6 a x log a 6 = 1. (63) 

Hence if the logarithm of a number to an}- base is known, 
its logarithm to any other base can easily be found. 



THE LOGARITHMIC SERIES. 141 

THE LOGARITHM CORRESPONDING TO A GIVEN 
NUMBER. 

73. We have heretofore regarded the logarithms as in- 
creasing uniformly, and the rate of increase of the numbers 
continually changing. We will now find a relation giving 
the increase in the logarithm when the numbers increase 
uniformly. 

Now equation (62) gives the value of b = e v , and hence 
putting nv for v, 

e nv _ yn _ J _|_ w .y _^ ^_ g^ c 

1 -2 1 .2-3 

which is the number whose logarithm is n. Denote this 

number by 1 + s ; then 

.9 = nv -\ h etc. 

1.21 .2-3 

Assume now that n = As -f JBs 2 + 6V + etc., and substitute 
this assumed value of n in the value of s ; and let us for 
the present, in squaring and cubing the assumed value of 
n, drop powers of s higher than the third. Equating the 
coefficients of the like powers of s, there is obtained 

Av = 1, Avhence A = - ; 

v 

Ah) 2 1 

Bv + — = 0, whence B = - — ; 

2 2v 

Cv + ABv 2 + — = 0, whence C= — ; 
G 3 v 

and the law of the series being seen, it is unnecessary to 
repeat the operation retaining higher powers of s. Hence, 



n = log (1 + s) = - s ■ 
v\_ 


_i 2 + S - 3 -^ + --etc.l. 
2 3 4 5 J 


Changing the sign of s, 




iog(i -,)=![- 


s 2 s 3 s 4 s 5 , ~| 

s etc. . 

2 3 4 5 



142 THEORY OF LOGARITHMS. 

Subtracting these two equations, and remembering that 
the subtraction of logarithms corresponds to a division of 
numbers, 

1 + s\ 2 



log 



s+~ + -+etc. 



, P — Q , 1 + s p 

N ow let s = — ; — ; whence h = — 

p + q' 1 — s q 



W E = 2 Sl^i + V^niY, IfP^lY +e tc 



.(64) 



If g = 1 and p = b, log 6 = 1. Substitute and solve for v : 

[HI 3^ + 1/5^ + 1, 
whence v can be computed for any positive base. 

Note that if the base is 1, v is zero ; and that if the base 
is less than 1, v is negative. 

Modulus. — The reciprocal of v is called the modulus of 
the system of logarithms. Now v has been shown to be the 
ratio of the rate of increase in the numbers in a system 
whose base is b (A at A , B at B^) to the rate of increase in 
the numbers in the natural system, the logarithms being 
conceived to increase uniformly. Hence the reciprocal 
of v, or the modulus of the system whose base is b, is the 
ratio of rate of increase in the logarithms in the system 
whose base is b (A at A , B at B^ to the rate of increase in 
the logarithms in the natural system, the numbers being 
conceived to increase uniformly. 

It will be seen from (64) that a logarithm in any system 
is inversely proportional to the value of v for that system. 
Now v is the reciprocal of M, and hence a logarithm in any 
system is directly proportional to the modulus of the sys- 
tem. The modulus of the natural system is 1 ; the logarithms 
of a number in any other system can therefore be obtained by 
multiplying the natural logarithm of the number by the modu- 
lus of the system. 



THE CURVE OF LOGARITHMS. 



143 



If in (64) M is written instead of - 
posed, we obtain 

\ogp = \ogq + 2M\ j^ 



q^3\p + q 



and logg is trans 
p — q \° lfp 



-fete. 



5 V p + q 

From wliieh the logarithm of any positive number can be 
found when the logarithm of any other number and also 
the modulus of the system are known. 

74. The logarithmic curve. — In the following diagram the 
abscissae of points in the curve represent numbers, and the 
corresponding ordinates represent the natural logarithms of 
these numbers. 
Y 




Fig. 81. 



75. Solve the following equations. — After obtaining the 
general solution, substitute 2 for a, 3 for, b 4 for c, 5 for d, 
6 for e, 7 for/; obtain the numerical answer in each case. 

130. a hx+c = d ex+f . ( a x+y = b, 

131. a hx c dx = e. ' \b x -y = a. 



132. ab x = cd x \ 



133. 



( a x = IP, 
( c x+d = e y+/ . 



135. a x % x c = 1. 

136. a h * = c. 



APPENDIX II. 



1. GONIOMETRY. 2. COMPLEX QUANTITIES. 
3. HYPERBOLIC FUNCTIONS. 

In the subject of Goniometry we will consider : (1) func- 
tions of the sum or difference of angles ; (2) the circular 
measure of an angle ; (3) arc functions. 

Review Chapter I. Functions of Angles. 

76. Functions of the sum or difference of angles. 

Denote the value of angle BAC by x, the value of angle 
CAD by y. From any point D 
in AD drop a perpendicular on 
AC and on AB. From C draw 
parallels to BA and DE, respec- 
tively. 

The angles CDF and BAC 
are equal because their sides 
are respectively perpendicular. 

AD sin (x + y) 
= ED = BC + FD. 

But BC = ACsinx 
and AC = AD cos y. 

Also FD = CD cos x and CD == AD sin y. 

Substituting : sin (x +/) = sin x cos/ + cos x sin/. 

Also, AD cos (x + y)=AE = AB- FC, 

AB = AC cos x = AD cos x cos y, 

FC = CD sin x = AD sin x sin y. 

Hence cos (x +/) = cos x cos/ — sin x sin/. 

144 




Fig. 82. 



(65) 



(66) 



SUM OR DIFFERENCE OF ANGLES. 145 

Had the figure been drawn with either x or y in any 
other quadrant than the first, the same result would have 
been obtained. 

Had y been subtracted from x, the point F would have 
been to the right of (7; FC would have been drawn from 
F to the left, and FD would have been drawn from F down- 
ward. 

Hence sin (x — y) = sin x cos y — cos x sin y. (67) 

Also cos (x — y) = cos x cos y + sin x sin _?/. (68) 

, . sin (x + y) sin x cos y + cos a; sin ?/ 

tan (a; + ?/) = ) '-4 = : r— ^« 

x y cos (a; -f- V) cos a; cos y — sm a; sm y 

Multiply numerator and denominator of this equation by 
sec x sec y. There is obtained 

tan x -f- tan y 



tan (x + y) — 



1 — tan x tan y 



~. ., , , N tan x — tan?/ 

Similarly, tan (a; — y) = T — - — — ■ 

J ' v * - J 1 + tan a ( tan y 

The other functions of (x ~f- ?/) and of (a? — ?/) can be ob- 
tained by taking the reciprocals of sin (x ± y), cos (x ± y), 
and tan (x ± y). 

By adding and subtracting (65) to (68), 

sin x cos y — l [sin (x + y) + sin (x — y)'], ■ (69) 

cos a? sin ?/ = i [sin (a; + ?/) — sin (a* — ?/)], (70) 

sin x sin y = — \ [cos (a- + ?/) — cos {x — y)~\, (71) 

cos a- cos y = \ [cos (x + ?/) 4- cos (x — y)]. (72) 

In these relations put \(A+B) for x and i(^4— i?) for 2/. 

sin A + sin 5 = 2 sin J (.4 + £) cos l (J. - 5), (73) 

sin .4 - sin B = 2 cos | (.1 + 5) sin 1 (.4 - 5), (74) 

cos .4 - cos B = - 2 sin i (J. + J5) sin 1 (^1 - 5), (75) 

cos .4 + cos B = 2 cos \ {A + -B) cos J (.1 - J5), (76) 



146 GONIOMETRY. 

TRIGONOMETRIC EQUATIONS. 

77. 1. Solve the equation a since -f b cos x = c. 

Put for cos x its value Vl — sin 2 x, and solve for sin x : 



ac±b Va 2 + b 2 - c 2 

sin x = — — 

a 2 + b 2 

Or proceed thus : Assume an angle y such that b=a tan?/. 
Substitute in the given equation and multiply by cos y : 

a (sin x cos y + cos x sin y) = c cos y. 

The solution is therefore obtained by obtaining y from 

the equation tan y = - and then finding x from the equation 
a 

sin (x + y)=- cos y. 

CL 

The first method of solution has the advantage of ex- 
pressing sin a; directly in terms of the given parts; the 
second method has the advantage of using formulae adapted 
to logarithmic computation. 

2. Solve the equation a tan x + b cot x = c. 
Multiply by tan x and arrange : 

a tan 2 x — c tan x + b = 0. 

ITrom which quadratic equation 



, c ± Vc 2 — Aab 

tan x = -= — 

2a 

Or proceed thus : from (11) and (9) 

1 — cos 2 x -, , 1 + cos 2 x 

tan a? = and cot x — — -, — 

sin 2 x sin 2 a? 

Substitute in the given equation and arrange: 
c sin 2 a? + (a — 6) cos 2 <c = a -f 6. 

If tan ?/ = , sm(2x + y) = cosy. 

c c 



TRIGONOMETRIC EQUATIONS. 147 

This solution gives a logarithmic solution of the general 
equation of the second degree. Thus in az 2 — cz + b = 
divide by z, and put tan x for z. 

The equation a tan x -f- b cot x= c is obtained. 

Hence z can be found by first finding y, then 2x-\-y, 
then x ; finally, z — tan x. 

3. Solve a (sin a; + tan x) ~b sin a? (esc x -\- cot aj). 
This equation can be arranged thus : 

(cos x + 1) (a sin x — b cos x) = 0, 

and the roots are obtainable by equating each factor sepa- 
rately to zero. 

4. Solve the equation a sin x + b sin (x -\- 10°) = c. 
By applying (65) and arranging, this becomes 

(a + b cos 10°) sin a; + b sin 10° cos x = c, 

which is of the form a sin x-\-b cos x = c. 

5. Solve the equation sin x + sin (a; — 10°) = .6. 

In (73), write x instead of A and x — 10° instead of B ; 
there is obtained 

sin x + sin (x — 10°) = 2 sin (a; — 5°) cos 5°. 

Comparing with the given equation, 

sin (x - 5°) = .3 sec 5° ; hence x = 22° 32' or 167° 28'. 

6. Solve the equation sin x sin (x — 10°) = .6. 

In (71), put x — 10° for y and compare the resulting equa- 
tion with the given equation 

cos (2 x - 10°) = cos 10° - 1.2 = -.2152, 

aj = 56° 13' or 133° 47'. 



148 GONIOMETRY. 

7. Solve the equation sin x = .6 sin (x — 10°). 
Apply (67), 

sin x = .6 sin a; cos 10° — .6 cos a* sin 10°. 

Multiply by esc 10° esc x ; there is obtained 

.6 cot x = .6 cot 10° - esc 10°. 
Prom which x = 135° 43' or 315° 43'. 

8. Solve the equation tan x tan (x — 10°) = .6. 

The quotient of (71) and (72) compared with the given 
equation gives 

cos (2 x - 10°) - cos 10° = 6 
cos (2x- 10°) + cos 10° ~~ * 

which gives cos (2 x — 10°) = .25 cos 10°, 

whence x = 42° 52' or 147° 8'. 

Problems. — Find the least positive angle satisfying each 
of the following equations : 

137. sin (10° + x) = .6 sin (20° + x). 

138. tan x = .6 tan (x — 10°). 

139. tan (x + 10°) 4 tan (x - 10°) = 2 cot x. 

140. sin (a 4 10°) 4 cos (x - 10°) = cos (x + 10°). 

141. cos x cos (x — 10°) = .6. 

142. 32 cos 2 ap + 3tan2a? + 3 cot (45° 4 x) = 0. 

143. sin (20° 4 *) cos (20° - x) = .6. 

144. sin 2x-\- sin 3 x = 3 sin a?. 

145. sin 2 a? — sin 2 (x — 70°) = .6. 

146. cot # tan 2x — tan xcot2x = 2. 

147. sin (c 4 a;) esc (c — x) = cos (a 4 b) sec (a — b). 



CIRCULAR MEASURE. 149 

78. The circular measure of an angle — Let A be the 

center of a circle, and CB q 

the arc included between 
the sides of angle A— 6.* 
With the same center 
A and any radius AB X 
draw an arc BjQ also in- 
cluded between the sides 
of A. FlG - 83 - B, 

The sectors ABC and AB X G X are similar, and hence 
arc BC : radius AB = arc B X C X : radius AB V 
That is, the ratio arc; radius is constant for any angle. 
This ratio is taken as the measure of angles ; its magnitude 
for any angle is called the circular measure of that angle. 
That is, in circular measure : 

, arc 

angle 




B, 



radius 

The arc of an angle generated by half a revolution (180°) 

is it times the radius. Hence 

angle of half a revolution = tr. 

The unit angle is obtained by dividing this equation by it. 

., i angle of half a revolution 
unit angle = — ^ 

This unit angle has been called a radian. Its value in 
degrees can be obtained from this last equation, 

1 radian = — = 57.296° = 57° 17'.75. 

7T 

If the angle $ in circular measure has x°, 

arc of angle degrees in arc of angle 9 







radius degrees in arc equal to radius 



That is, 0= ± - and a = 57.296 0. 

57.296° 

* Angles measured in radians will have their values denoted by 
Greek letters, w (Pi), d (Theta), <p (Phi), etc. 



150 



GONIOMETBY. 



A table for converting degrees into radians, or for con- 
verting radians into degrees, will be found at the bottom 
of page vii of the tables. 

Exercises. — The following table gives the values of 
various angles, first, in degrees and minutes ; second, in 
radians ; third, in terms of an angle of half a revolution : 



A>~GLE IX 


A>'GLE IX 


ANGLE IN 


Angle in 


Angle i>- 


Angle in 


Degrees. 


Radians. 


Terms of it. 


Degrees. 


Radians. 


Terms of it. 


172° 


3.0020 


.9556 


144° 9' 


2.5133 


.8008 


100° 


1.7453 


.0006 


169° 23' 


2.9563 


.9401 


125° 


2.1566 


.6944 


104° 51' 


1.8300 


.5825 


179° 


3.1241 


.9944 


204° 12' 


3.5639 


1.1344 


194° 


3.3859 


1.0778 


195° 33' 


3.4130 


1.0864 


300° 


5.2360 


1.6667 


274° 10' 


4.7851 


1.5231 


344° 


6.0039 


1.9111 


289° 50' 


5.0585 


1.6102 


354° 


6.1784 


1.9667 


344° 19' 


6.3355 


2.9667 



79. Arc functions. — The equation X=sin6 gives the 
value of X as a function of 6 ; it is frequently convenient 
to express the value of 6 as a function of X. The notation 
employed is : 6 = arc sin X* This will be read, 6 is the 
angle (or arc) whose sine is 3". Similarly = arc tan N is 
read, 6 is the angle (or arc) whose tangent is X; and simi- 
larly for each of the other functions. 

It will be seen that if we omit the name of a trigono- 
metric function prefixed to one side of an equation, the 
name of the corresponding arc function is prefixed to the 
other side of the equation ;. and reversely. 

It must be remembered, however, that in the equation 
X= sm 6, there is but one value for J\ r for each value of 0; 



* By some authors this is -written, = sin -1 X. 



ARC FUNCTIONS. 151 

but in the equation = arc sin JST, there are an infinite num- 
ber of values of for any value of the number JV. Thus 
in the equation JV= sin 0, if = Jtt or 30°, N= .5; but in 
the equation = arc sin JV) if 2V= .5, = [?i -f- ( — l) 71 -^]^, 
n being any whole number. 

Exercises. — 1. Find 6 = arc tan 1. 
Here tan 9 = 1, and = (n + £)7r. 

2. Find JV= tan (arc cos f). 

Let arc cos £ = 0, then cos = \\ = (2n ± -§-)tt. 
Hence iV= tan (2w ± £> = ± 1.7321. 

3. Prove that sin(arc sin JV) = JV, 
and that arc sin (sin 0) = 0. 

2 iV 

4. Prove that 2 arc tan JV= arc tan • 

1 — N 2 

5. Arc sin 2F+ arc sin 1 ^= | ?r. Find JV. 

6. Arc sin 2 N— arc sin .A 7 V3 = arc sin N. Find JV. 

PROBLEMS IN GONIOMETRY. 

148. At a distance of 70 feet from the foot of a tower, 
the angle of elevation of the top is 21° greater than the 
angle of elevation seen at a distance of 160 feet from the 
foot of the tower. Find the height of the tower. 

149. At a distance of 296 feet (a) from the base of a 
tower, the angle of elevation of the top is three times as 
great as the angle of elevation seen from a point 208 feet 
(b) nearer the base. Find the height of the tower. 

150. The wind strikes a horseman at an angle of 50° with 
the road when he is traveling at the rate of 8 miles an hour ; 
and at an angle of 40° when his rate is 12 miles an hour. 
Find the direction and velocity of the wind. 



152 COMPLEX QUANTITIES. 

151. Prom the top of a hill 360 feet above the surround- 
ing plain, an officer sees a column of the enemy's infantry 
advancing directly toward the hill. The column is known 
to be one mile long, and at a certain time the whole column 
is seen by the officer to subtend an angle of 2° 7'. At that 
time what is the distance of the front rank from the middle 
of the base of the hill ? 

152. A circle has a radius of 20 inches (?•). Froni the 
same point in the circumference two chords are drawn of 
lengths 10 inches (A^) and 15 inches (AV), respectively. Find 
the area of the portion of the circle bounded by these chords 
and by the arc joining the ends of the chords. (Two solu- 
tions.) 

153. A circular field, surrounded by a fence, contains 1 
acre. A cow is fastened to the fence by a chain, which, 
when measured, was found to be 55.53 feet long. Over 
what area can the cow graze inside the circle ? 

COMPLEX QUANTITIES. 

80. A complex quantity is of the form a + &*, where i 
stands for ^ — 1 ; a is called the real part of the complex 
quantity, and hi is called the imaginary part. 

It has already been explained that the multiplication of 
a segment by — 1 reverses the direction of the segment. 
This reversal may be conceived as being made by means of 
two half-reversals. Since i 2 = — 1, i will be taken as the 
multiplier which first turns OU (Fig- 84) into OU lt and then 
turns OUi into OU 2 \ multiplying again by i. OU 2 is turned 
into OU s ] and another multiplication by i turns OU 3 into 
OU 

Real positive quantities have been conceived as laid hori- 
zontally to the right, and real negative quantities to the 
left. Positive imaginary quantities will be conceived as 



ARGUMENT AND MODULUS. 



153 



being laid vertically upward, and negative imaginaries ver- 
tically downward. 

Thus the complex quantity a -+- bi will be represented by 
laying off OM = a to the right of the origin and MA 1 = b 
upward from M. If a is negative it is laid off to the left 
of y and if b is negative it is laid off downward. 



A, 


Uj 1 




i 


u, 




y^ a 


A 






-y 


/t> 


-/ 


+/ 


Ju 





A, 








uT 






A 



M 



Fig. 84. 



Denote the angle MOA x by and the length of the line 
0A l by m. Then a = b tan = m cos 9 and b — m sin 0. 

Hence the complex quantity a + bi may be written : 
m (cos + i sin 0). The angle 6 is called the argument of 
the complex quantity, and jra is called its modulus. 

It will be noticed that the segment OU (+ 1) is changed 
into the segment OA x by two distinct operations. 

First. A turning of OU through the argument 6 into 
Q [ This turning is therefore equivalent to a multiplica- 
tion of + 1 by cos 6 + i sin 0. 

Second. A multiplication of the length of OU a by the 
modulus m ; the positive square root of a 2 + b 2 . 



154 



COMPLEX QUANTITIES. 




Fig. 85. 



Addition of complex quantities. — Suppose it is desired to 
add the complex quantities m 1 (cos 9 + i sin 6) and m 2 (cos <£ 
-f-isin<£). Make angle MOA—0, OA—m^^ angle NAB=cf>, 
AB=m 2 ; the required sum is OA+AB. The sum of the real 
parts is m l cos -+- m 2 cos <f> 
(OM+AN); and the sum 
of the imaginary parts is 
i (m 1 sin -\-m 2 sin <f>) ; that is, 
MA + tfB. 

Had ra 2 (cos <£ + i sin <£) 
= OC been laid off first and 
m x (cos <£ -f- i sin <£) = (75 been 
added to it, the result would 
have been precisely the same. 
Hence the segment OB rep- 
resents the sum both in mag- 
nitude and direction. 

A third complex quantity can be added to this sum by 
drawing from B a segment, say BD, representing in magni- 
tude and direction this third complex quantity. Then the 
segment OD will represent 
both in magnitude and direc- 
tion the sum of the three com- 
plex quantities, and so on. 

Subtraction. — To con- 
struct the difference of 
m 2 (cos <£ + ?' sin <f>) 
and m 1 (cos -+- i sin 0), 
lay off the minuend (OA); 
let AB represent the sub- 
trahend; lay it off back- 
ward from AB to the point 
B l ; then 0B 1 will represent the remainder both in magni- 
tude and direction. 




Fig. 86. 



FUNDAMENTAL OPERATIONS. 



155 



Multiplication, — The product of the two complex quanti- 
ties mj (cos + i sin 0) and m 2 (cos <f> + i sin <£) is 

m x m 2 [(cos cos <f> — sin sin <£) + i (sin cos <£ + cos sin <£)] 
= ra^ [cos (0 + <£) + t sin (0 + <£)], (77) 

the arguments being added and the moduli multiplied. 

Just as OU (+1) was 
turned through an angle 
and then multiplied by m 
when operated upon by the 
multiplier 

m (cos + i sin 0), 
so OA is turned through 
an angle (*70J3 or AOC) 
and then multiplied by m 2 
when operated upon by the 
multiplier 

ra 2 (cos <£ + i sin <£). 

If (Fig. 87) angle *7(L4 = 
0, UOB = <f>, U0C=6 + <I>, then BOC=UOA. Also the 
length 0(7 of the modulus = m^ is a fourth proportional 
to 1, ra 2 , and m x ; that is, OU: OB = OA: OC. Hence the 
two triangles UOA and BOC are similar. 

To construct OC, therefore, it is sufficient to make angle 
BOO =6 and angle B = angle OUA. 

If now OC is operated upon by a new multiplier, 

ra 3 (cos \f/ + i sin i/>), 

the new argument will be + <£ + i/a, and the new modulus 
will be m^wig. The geometric construction can be obtained 
as before. 

Division. — Kationalizing the denominator of the fraction 

wii (cos + i sin 0) 
??i 2 (cos <f>-\-i sin <£) 




Fig. 87. 



156 

gives 



m 2 



COMPLEX QUANTITIES. 



-(cos 6 -f- i sin 0)(cos <f> — i sin <£) 



»i 



= — - [cos + i sin 0][cos(— <£)-h i sin(— <£)] 



m 



= 5l [cos (0 - <£) + i sin (6> - <£)], 

9H>2 

the arguments being subtracted and the moduli divided. 




To construct the quotient OC, make angle AOC '= <f> and 
angle A = B. Then 

the argument UOC= UOA - {705 = — <}>; 

and since OC : 0*7= 0^1 : OB, the length of 0(7 = 

the modulus of the quotient = — -• 

m 2 

Powers and roots. — The factor ??i(cos0 + i sin 0) repeated 

n times gives by (77) : 

\_ni (cos + i sin 0)] n = m n (cos ?i0 + i sin ?i0). (78) 

Write - instead of 6, m instead of m n , and take the nth 
n 

root 



m n ( cos - + » sin - ^ = [??i (cos + i sin 0)1". 

\ ft n; 



POWERS AND ROOTS. 



157 




Fig. 89. 



To construct the power, lay off, upward from the half-line 
U, n angles each equal to UOAy = 0. 

Make OA x = m and join UA V v \ 
Make angles OA x A 2 , OA 2 A s , etc., N N 

each equal to OUA v Then OU=l, 
OA x = m, OA 2 = m 2 , OA 3 — m 3 , etc. 

Now the arguments UOAy, UOA 2 , 
UOA 3 , etc., increase in arithmetical 
progression ; the segments OA x , OA 2 , 
OA 3 , etc., which represent the com- 
plex quantities m (cos 6 -f- i sin 6), 
m 2 (cos 0+ i sin 8) 2 , m 3 (cos 0-f i sin &f, 
etc., increase in geometrical pro- 
gression. 

It follows that the arguments 6, 
2 0, SO, etc., are the logarithms of 

the corresponding complex quantities. Now denote m by 
k e and take the Oth. root of k (cos + i sin 0). This root is 
k (cos 1 + i sin 1), which is the number whose logarithm is 1, 
and therefore the base of this system of logarithms. 

The curve JJA Y A 2 A Z (Fig. 
89) is called the logarithmic 
spiral. It will be seen that 
it may be used as a graphical 
table of logarithms ; the 
same thing is true of the 
curve in Fig. 81. 

When m = 1, the points 
Ay, A 2 , A 3 , etc., will be ar- 
ranged at equal distances 
from each other on a circle 
of unit radius, Fig. 90. In 
this case k also equals 1 and the base of the system of 
logarithms is cos 1 -f- i sin 1 — .5403 + .8415 i. 




Fig. 90. 



158 



COMPLEX QUANTITIES. 



Construction of roots. — It has 

already been shown that the nth 
root of m (cos 6 -h i sin 6) was 



,nU 



mn( cos- + i sin- ,, 
n n, 



where m is the positive square 

i 
root of a 2 -f- b 2 ; m» will here de- 
note the real positive nth root of 
m. Now the functions of 2sir 
4- are the same as those of 0, 
and hence 



m (cos -f- i sin 0) h = m» 



cos 




and by taking values of s = 0, 1, 2, 3, etc., the n roots are 

2 7T 



.»[ cos- + i sm- 



V 2tt + , 

m»( cos ■ — + i sin 

71 






If 4tt + , . • 4tt + 0\ , 

m«( cos — — h i sin =! — - ], etc. 

V n n J 

Taking $ = -^ir = 50°, m = 3.052, n = 5, the roots are 
1.25 (cos 10° + isin 10°), 1.25 (cos 82° + i sin 82°), 
1.25 (cos 154° 4- i sin 154°), 1.25 (cos 226° + i sin 226°), 
1.25 (cos 298° -M sin 298°), 
giving the points R x , R 2 , i? 3 , R 4 , R 5 . 

By taking m = l and £=0, the n roots of +1 are obtained. 
By taking m=l and x=tt, the n roots of —1 are obtained. 
By taking m=l and x=%7r, the n roots of % are obtained. 
By taking m = l and »=§ ir, the n roots of — i are obtained. 



SERIES FOR SIN AND COS 0. 159 

81. Values of functions of multiple angles. — Put 1 for m 

in (78) : cos nO + i sin nO = (cos -f i sin 0) n . (79) 

Expand by the binomial formula ; equate the imaginary 
parts and also the real parts : 

sin nO = n cos n_1 sin — n \ n ~ J^ n ~ ' cos n_3 sin 3 + etc. 
cos nO = cos n — n \'~ ' cos n ~ 2 sin 2 + etc. 

JL • Zi 

Thus to obtain the functions of 3 0, make n = 3, 

sin 3 = 3 cos 2 sin (9 - sin 3 = 3 sin - 4 sin 3 0, (80) 
cos 3 = cos 3 - 3 cos sin 2 = 4 cos 3 - 3 cos 0. 

82. Exponential value of cos 8 + / sin 9. — When is zero, 
cos$-Msin0is 1. Now let us imagine a point generating 
a circle ; when the angle is zero, counted from a horizontal 
initial line, the horizontal component of the motion of the 
generating point is zero, and the vertical component of its 
motion is equal to the motion in the arc. Therefore when 
is at zero, the rate of increase of cos -f i sin is i times 
the rate of the generating point. If we assume the gener- 
ating point to move a unit distance in unit time, the rate of 
increase (0 = zero) of cos -f- 1 sin is i. 

Now is the logarithm of cos + i sin ; therefore, by 
the theory of logarithms, cos + i sin = e 6v , v being the 
rate of increase in the number cos -f- i sin 0, when is 
zero and increasing at unit rate. 

Therefore cos + i sin = e 6i • 

Changing the sign of : cos — i sin = e~ m ; 

whence sin = — (e ei — e~ ei ) 

= e 1- + —1 t + ctc. (81) 

1 1.2-3 1.2.3.4.5 1.2.3.4.5.6.7 V ; 

cos0 = ±(e ei + e~ ei ) 

= 1 £- + 1 1 + e tc. (82) 

1.2^1.2.3.4 1.2.3.4.5.6^ V ; 



160 



HYPERBOLIC FUNCTIONS. 



HYPERBOLIC FUNCTIONS. 

83. Definition of an equilateral hyperbola. — Suppose a 
double cone (called a cone of two nappes) with vertex at 
V. If this cone is cut 
through the axis, two 
angles, BVE and DVQ, 
will be formed. If the 
cone is such that this 
angle is 90°, and the 
cone is again cut by a 
plane OA normal to the 
plane BVE, this new 
section is bounded by 
a curve of two branches 
QAB and Q^i^; this 
curve is called an equi- 
lateral hyperbola. The 
points A and A 1 are 
called the vertices of the 
hyperbola, the point 0, 
midway between them, 
is called the center of 
the hyperbola, and AA X 
is called the axis of the 
hyperbola. 

The two following 
properties of the equi- 
lateral hyperbola are 
proved in treatises deal- 
ing with this curve. 

First. If Q is a point on the curve, and O^is the abscissa 
of ft 

ON 2 - NQ 2 = OA 2 . 




Fig. 92. 




Fig. 93. 



THE EQUILATERAL HYPERBOLA. 



161 



Second. Area segment AQN of the hyperbola 

= i ON x NQ - x OA> log, f **^®)-* 
Now the sector OJ. Q == triangle ONQ — segment AQN. 
Hence sector O^IQ = \ OA 2 log e / O^+W j. 

The quantity log/ 0iV "+ xY ^ will be denoted by u; and 
the semi-axis (X4 will be denoted by a. 
Hence Sector OAQ = \a*u. 

84. Definitions of the hyperbolic functions. — Let be the 

center of an equilateral hyperbola, OA(=a) its semi-axis, 
Q any point on the curve, 
ON the abscissa of Q, and 
NQ the ordinate. With a 
center and a radius a, 
describe a circumference ; 
let P be a point on the cir- 
cumference ; let the length 
of arc AP be equal to ad ; 
then 6 represents the cir- 
cular measure of the angle 
AOP. The area of the cir- Fia 94 ' 

cular sector AOP equals its arc times \ the radius = ±a 2 0. 

* 1. The equation of the two generators of the cone are x— y=0 and 
x + y = 0, their product being x 2 — y 2 =0. As these generators revolve 
around the axis of X, x remains constant and y 2 becomes y' 2 + z 1 . 
Hence the equation of the double cone is x 2 — y 2 — z 2 = 0. If now the 
cone is cut by the plane z = «, the projection of the section on the 
xy-plane is x 2 — y 2 = a 2 . 

2. The integral of ydx or Vx' 2 — a 2 dx is 




| xy/x 2 — a 2 — \ a 2 log e (x + Vx' 2 — a 2 ) . 
Taking x for the upper limit and a for the lower, and putting y for 
Vx 2 - a 2 , the area is \ xy - \ a 2 log e {^l±v\ . 



162 HYPERBOLIC FUNCTIONS. 

From P, drop a perpendicular PM on OA. Then PM 

PM 

= a sin 0, OM— a cos 6, and — — = tan 0, and so on. From 

OM 

these properties of PM and Oilf, the trigonometric functions 
have been also called circular functions. 

From analogy, certain ratios of the semi-axis, the abscissa, 
and ordinate of a point on an equilateral hyperbola are 
called hyperbolic functions. 

They are denned as follows : Let u be such a quantity 
that 

Area sector OAQ = \a 2 u. Then 

1. The ratio of NQ to OA is called the hyperbolic sine 
of u. Hyperbolic sine u will be written sinh u and will be 
read " shin it." It follows that QN = a sinh u. 

2. The ratio of ON to OA is called the hyperbolic cosine 
of u ; written and pronounced cosh u. It follows that 
OiV=acosh^. 

3. The ratio of sinh u to cosh u (NQ : ON) is the hyperbolic 
tangent of u (tanh u). 

4. The reciprocal of cosh u (OA: ON) is the hyperbolic 
secant of u (sech u). 

5. The reciprocal of sinh u (NQ : OA) is the hyperbolic 
cosecant of u (csch u). 

6. The reciprocal of tanh u (ON: NQ) is the hyperbolic 
cotangent of u (coth u). 

85. Exponential values of hyperbolic functions. — By the 

property of the equilateral hyperbola, 

ON 2 - NQ 2 = OA 2 . 
Dividing by OA 2 , cosh 2 u — sinh 2 u — 1. 
But since „ = log„(M±_W 

cosh u + sinh u = e u . 
Hence, by division, 

cosh u — sinh u = e~ u . 



RELATIONS OF THE FUNCTIONS. 163 

Therefore, by subtraction and addition, 

sinh u = \ (e u - er u ), (83) 

cosh u = i (e M + e~ w ). (84) 

GENERAL RELATIONS. 

86. Relations of hyperbolic and circular functions. 

1 



Now \{f —e~ u ) = — i 



m (e uii — e~ uii ) 



Hence by (81) and (83) : sinh u = — i sin ui. (85) 

Also, i (e u + e~ u ) = \ (e uii + e~ uii ). 

Hence by (82) and (84) : cosh u = cos ui. (86) 

Periodicity of hyperbolic functions. 

Now — i sin ui = — i sin (m — 2 mi) 

= — i sin (w -f 2 ?i7ri) ^. 
By (85), sinh u = sinh (it + 2 nTrt). (87) 

Similarly, cosh u = cosh (w + 2 W7r?T). (88) 

The addition formulae. — From (85), (86), (65), and (66) : 
sinh (u -f v) = — i sin (ui + vi) 

= — i (sin ui cos vz" + cos ui sin m) 

= sinh u cosh v -f cosh u sinh v. (89) 
cosh (u +v) = cos (iu+m) = cos ui cos vi—smui sinvi 

= cosh u cosh ?; + sinh u sinh v. (90) 

Functions of multiples of u. — From (89) and (90) there 
is obtained 

sinh 2u = 2, sinh u cosh u, 

cosh 2u = cosh 2 u + sinh 2 w, 

sinh 3 u = 4 sinh 3 w -f- 3 sinh u, (91) 

cosh 3 w = 4 cosh 3 u — 3 cosh ?z. (92) 




164 HYPERBOLIC FUNCTIONS. 

87. Longitude of u. — With, center N and radius equal to 
NQ, draw an arc intersect- v Q 
ing the circle in P. Join 
NP and OP. Since J 

ON 2 - NQ 2 = a 2 , 
ON 2 -NP 2 = OP 2 . 

Therefore, the angle OPN 
is a right angle. Denote 
the angle PON by x. Fig. 95. 

sinh u = NQ : OA=NP:OP= tan x, (93) 

cosh u=ON: OA = ON:OP= sec x. (94) 

In (11) put 90° + x for x ; then 

tan a; + sec x = tan (45° + ^x). 
Therefore tan (45° +±x)= sinh u -f cosh w = e u . 
Hence log tan (45° + \x)=u\oge = .43429 w. 
From which for any value of u, x can be found, and thence 
in (93) and (94), sinh u and cosh u can be found. 

The angle x was called by G-udermann the longitude of u. 
Some later writers, however, have referred to it as the 
G-udermannian function of u. 

A table at the bottom of page vii gives values of u corre- 
sponding to values of x, the longitude of u, within limits 
which admit of interpolation. 

88. Solution of cubic equations. — Consider the cubic equa- 
tion w z ± aw = b. 

First. The coefficient of w is positive. 

Let w = 2smhuJ-. (95) 

The given equation reduces to 

sinh 3 u -f- § sinh u = — \— 
Sa *a 

Hence by (91), sinh 3 u = |* ^ (96) 



CUBIC EQUATIONS. 165 

Denote three values obtained from this equation by 3 u, 
3 u + 2 iri, 3u — 2 iri. 
By (89), 

sinh (u ± | iri) = sinh u cosh -| iri ± cosh u sinh |- 7ri 
= sinh u cos 120° ± i cosh tt sin 120°. 
Hence, to solve the equation under consideration, find u 
from (96) ; then by (95) the three roots of iv are 

2 sinh u -v/-, and -%/- (— sinh it ± iV3 cosh w). 

Second. Consider the equation w s — aw = b. 
Case I. 27 6 2 is greater than 4 a 3 . 

Let w = 2 cosh w ■%/-■ 



The given equation reduces to 



36 IS 

8a ^a 



cosh 3 u — f cosh 

Hence by (92), cosh 3 u = p ^- (97) 

Find u from (97) ; then the three roots are 

2 cosh u a|-, and -*/- (— cosh u ± iV3 sinh u). 

Case II. 27 6 2 is less than 4 a 3 . — This is the irreducible 
case of Tartaglia's solution, in which all the roots are real. 

Let w = 2 sin x -%/— 

The given equation reduces to 

• o , • 3 6/3 

snr x — f sin # = — \ — 

8a *a 



Hence by (80), sin 3 x = - 1^ J?. 

2 a *a 

Find a; from (98) ; then the three roots of iv are 
2sina^/| 2 sin (a; ±120°)^. 



(98) 



166 HYPERBOLIC FUNCTIONS. 

Exercises. — 1. Find the value of i\ 

Raise to the i power the equation cos 6 + i sin 6 = e 9i . 

(cos 6 + i sin Of = e~ 9 . 

Make = (2 » + \) ir. 

i" =z= e-^ x e 2n7r = .2079 e 2 ™, 

where n is any whole number ; when n is zero, £ = .2079. 

2. Solve by Horner's Method the equation 

ar*-fa; + isiii30 o = 3 
and thus find sin 10°, sin 50°, and sin 130°. 

3. By substituting — for 6 in (81), find sin 10°. 

18 

4. Solve the following cubic equations : 

x 3 + 24.x = 511 x 3 - 6x = 9 x 3 - 9 a; = 10 

a* + 54 a; = 288 X s - 7x = 90 ar 3 - 7a; = 6 

ar 5 + 72 x = 400 ar 3 - 48 x = 520 ar 3 - 39 x = 70 



PROBLEMS FOR EXAMINATION. 



[No answers are given for these problems.] 



RIGHT TRIANGLES. 

154. A person standing at a distance of 143 feet from the 
middle of the base of a tower finds the altitude of the tower 
to be 43° 10'. Find the height of the tower, the eye of the 
observer being 5 feet from the ground. 

155. From the top of a lighthouse 127 feet above the 
level of the sea, the angle of depression of a boat was seen 
to be 10° 20'. Find the distance of the boat from the foot 
of the lighthouse. 

156. At what angle do we ascend a regular acclivity 6.5 
miles long, attaining an altitude at the summit of 4268 feet ? 

157. The horizontal distance between two columns is 124 
feet, and the straight line joining the tops of the columns 
makes an angle of 12° 40' with the horizon. Find the dis- 
tance between the tops. 

158. A canal boat is drawn along the middle of a straight 
canal by two men hauling upon ropes 60 feet and 70 feet 
long, respectively. Find the ratio of the strains upon them, 
the canal being 20 feet wide. 

159. A row-boat, pulled at the rate of 5 miles an hour, is 
required to cross a river 1.3 miles wide to a poi-nt which is 
.78 mile lower down the stream than the point of starting. 
What angle must the direction of the boat's head make with 
the bank, supposing the stream to run at the rate of 4 miles 
an hour ? 

167 



168 PKOBLEMS FOB, EXAMINATION. 

160. If d is the distance (in degrees) of the sun from 
the equator, show that the altitude of the sun at midday is 
90° -\- d — l, where I is the latitude of the place. Hence, 
find the length of the shadow at midday of a tower 120 feet 
high, the latitude of the tower being 40°, and the sun being 
vertically above a point in 10° north latitude. 

161. At midday of the 21st of March, the length of the 
shadow of a 12-foot pole was 10.6 feet. What was the lati- 
tude of the place ? 

162. The problem of Eratosthenes. At midday, when the 
sun was vertically over Syene, the shadow of a vertical rod 
at Alexandria was measured, the shadow being found to be 
.13 times the length of the rod. Taking Alexandria to be 
5000 stadia due north of Syene, find the circumference of 
the earth. 

163. The latitude of Berlin is 52° 30'. Taking the earth 
to be a sphere whose radius = 3960 miles, find the radius 
and length of 1° of the parallel passing through Berlin. 

164. What will be the visual angle of a sphere 6 feet in 
diameter, if the eye of the observer is placed 7 feet from its 
center ? 

165. On a hill over the sea is a 60 feet high tower. From 
a boat, the angle of elevation of the foot of the tower is 14°, 
and of the top of the tower is 20°. Find the height of the 
hill. 

166. A flagpole, 20 feet high, is on the top of a tower. 
At a horizontal distance of 420 feet from the foot of the 
tower, the angle of elevation of the highest point of the 
flagpole is seen to be 15° 20 '. Find the height of the tower. 

167. The lune of Hippocrates. A semi-circumference and 
a quadrant concave in the same direction are drawn to meet 
at their extremities so as to inclose space ; the radius of the 
semi-circumference is 10 feet : find the area inclosed. 



RIGHT TRIANGLES. 169 

168. On one side of an angle A — 50°, a distance AB = 10 
feet is taken. From B, a perpendicular BC is let fall upon 
the other side of the angle ; from C, a perpendicular CD is 
let fall upon the first side ; from D a perpendicular DE is 
let fall upon the second side ; and so on, ad infinitum. Find 
the sum of all the perpendiculars. 

Put A instead of 50° and a instead of 10 feet ; then obtain 
the general solution of this problem.* 

169. A circle has a radius of 22 inches. From the same 
point in the circumference, two chords are drawn of lengths 
10 inches and 17 inches respectively. What angle is formed 
by these chords ? 

170. Two towers stand 100 feet apart on the same hori- 
zontal plane. An observer in line with them, and 150 feet 
from the nearer one, sees the angle of elevation of both of 
them to be the same. He then walks directly toward them 
a distance of 130 feet, when he finds the angle of elevation 
of the nearer one to be twice as great as that of the farther 
one. Find the heights of the towers. 

Before performing the computation for this problem, 
obtain the general solution of it. 

171. The top of a hill is 90 feet higher than the foot of a 
tower ; from the top of the hill the angle of depression of 
the top of the tower is 5°43-j-'. The horizontal distance 
from the tower to the hill is 399 feet. Find the height of 
the tower. 

172. I bought a lot having the shape of a right triangle ; 
the hypotenuse = 875 feet, and one of the acute angles = 
37° 53'. How much was paid for the land if the price was 
2.7 cents per square foot ? 

* If for the data of a problem, literal values are assumed instead of 
numerical values, the answer to the problem expressed in terms of 
these literal values is called the general solution of the problem. 



170 PROBLEMS FOE EXAMINATION 



OBLIQUE TRL1NGLES. 

173. To measure the distance between two points A and 
B separated by a pond, a third point was chosen. AC 
was found to be 545 feet. CB = 305 feet, and the angle 
ACB = 67=. Find AB. 

174. From a point on the bank of a river, an observer 
goes directly away from the river 165 yards up a slope 
inclined at an angle of 12° 12' to the horizontal; at this 
point the angle of depression of an object on the opposite 
bank and in the same plane with the first position was 
found to be 2° 10'. Find the breadth of the river. 

175. TJit range finder. From a fort a torpedo-boat is 
observed to be exactly east: eighty seconds afterward 
it is exactly northeast, and eighty seconds after that it is 
N. 22° 30' E. The torpedo-boat is going at a uniform rate 
in a straight line. Find its bearing at the end of the next 
eighty seconds. 

176. A vertical tower stands on a plane inclined 11 : 20' 
to the horizontal. Pleasuring from the base of the tower 
down the slope an oblique distance of 59.18 feet, the tower 
was found to subtend an angle of 38°. Find the height of 
the tower. 

177. Two inaccessible points X and Y are observed from 
v" : ~ ": : i . B. 500 feet apart (A and X on the left). 
The angle XAY= 42=. angle YAB = 67°, angle ABX = 39°, 
angle XBY= 47 : . Find XY. 

178. A house is seen from two ships A and B which are 
300 feet apart. At A. the angle of elevation of the house 
is found to be 28° 20' and the horizontal angle from it to the 
ship B = 48 = 40'. At B. the horizontal angle from the house 
to the ship A is found to be 2t"> : 15'. Find the height of the 
house above the level of the water. 



OBLIQUE TRIANGLES. 171 

179. To find the length BC of a lake running due north 
and south (C north of B), a point D was taken 530 feet 
north of C and a point A was found due west of B. It was 
found that AC = 1970 feet and AD = 2400 feet. Find the 
length of the lake. 

180. To find the distance between two inaccessible points 
A and B, an observer chose a point C in the same plane and 
found angle ACB = 113°. Then he walked on the continua- 
tion of AC to a point 125 yards from C, where the angle 
subtended by AB was found to be one-half of angle ACB. 
Then he walked on the continuation of BC 173 yards from 
C and again found the angle subtended by AB to be one-half 
of angle ACB. Find AB. 

181. Given the three sides of a triangle to find the length 
from the vertex to a point of trisection of the base. Take 
the base (c) = 200 feet, a = 205, b = 85, and find the length 
to the point of trisection near a. 

182. How many square miles of the earth's surface can 
an observer see if stationed at the top of Mt. Everest, whose 
altitude is 29,000 feet ? The radius of the earth = 3960 
miles. 

183. A lighthouse 60 feet high is on a cliff over the sea. 
From the top of the lighthouse the angle of depression of a 
boat was seen to be 16° 20' ; and from the bottom of the 
lighthouse, the angle of depression of the boat was 12° 10'. 
Find the height of the cliff and the distance of the ship 
from the foot of the cliff. 

184. Wishing to know the breadth of a river from a point 
X to a point A, an observer measures in the prolongation of 
XA a distance AB = 500 feet ; then from B, he measures 
at right angles to XAB a distance BC = 700 feet ; from the 
point C, the angle XCA was found to be 24° 20'. Find the 
breadth of the river. 



172 PROBLEMS FOR EXAMINATION. 

185. P, Q, and R are three known points in a straight 
line ; PQ = 104, QR = 296 ; PQ and QR are observed to 
subtend equal angles = 46° 50 ' at a certain point S. Find 

SQ. 

186. From the bottom of a wall 20 feet high, the angle 
of elevation of the top of a house = 40° 20' ; at the top of 
the wall the angle of elevation of the same is 19° 40'. Find 
the height of the house. 

187. A person at a point A, wishing to know the distance 
of an inaccessible object C on the opposite side of a river, 
lays off a base AB = 898.8 ; he then measures the angle 
CAB = 63° 20' and angle CBA = 52° 40'. Find AC. 

188. Of three towns the second is 11.6 miles from the 
first, the third is 10.5 miles from the second, and the first is 
14.3 miles from the third. Find the angles of the triangle 
whose vertices are the three towns. 

189. From the deck of a ship, 10 feet above the sea, the 
angle of elevation of the top of a cliff is 20° 20' ; from the 
top of the mast 65 feet higher, the angle is 12° 30'. Find 
the height of the cliff from the sea and the distance of the 
ship from the foot of the cliff. 

190. From the top of a tower 60 feet high, a column sub- 
tends an angle of 28° ; from the base of the tower, the col- 
umn subtends an angle of 40°. Find the height of the 
column. 

191. A balloon is ascending vertically with a uniform 
velocity. When it is one mile high, its angle of elevation is 
32° ; twenty minutes later, the elevation is 54°. How fast 
is the balloon rising ? 

192. At two points, 70 and 50 feet from the foot of a 
tower, a flagstaff on the top of the tower subtends the same 

, Find the length of the flagstaff. 



REDUCTION TO FIRST QUADRANT. 173 

FUNCTIONS OF ANGLES. 

Find a function of an angle of the first quadrant equiva- 
lent to each of the following functions : 

193. sin 283°. 201. sin 324°. 209. sec 247° 42'. 

194. cot 196°. 202. esc 562°. 210. esc 176° 28'. 

195. cos 344°. 203. cos 897°. 211. cot 31 2° 53'. 

196. sec 127°. 204. tan 631°. 212. sec 228° 12'. 

197. tan 274°. 205. tan 189° 49'. 213. tan 142° 29'. 

198. cos 245°. 206. cos 128° 11'. 214. cos 218° 47'. 

199. esc 144°. 207. sin 212° 14'. 215. sin 374° 55'. 

200. cot 133°. 208. cot 321° 37'. 216. esc 524° 19'. 

Eeduce each of the following expressions to its simplest 
form : 

217. a cos (180° - x) + b cos (90° + x) cot x. 

218. k sin x cot x tan (180° + x) sec (90° — x). 

219. tan (180° - x) cot (270° - x) + cot (90° + x) cot x. 

220. cos (90° + x) sin (360° - x) - sin (270° - x) cos (- x). 

221. (a + b) 2 sec (360° -x)-(a- b) 2 esc (90° + x). 

222. sec (90° + x) esc (360° - x) + tan (90° + x) cot x. 

223. tan x cot x + tan (— y) -j- tan (180° -f- y). 

224. sin (180° - x) cos (90° + x) esc (180° + x) esc x. 

225. (a 2 + b 2 ) sec (270° - x) + (a 2 - b 2 ) esc (180° + x). 

226. a sin (360° - a?) tan (270° + x) sec (180° - x). 

227. 2 cot 2 x cos x + 2 sin x — esc x. 

228. a 2 sin 90° + 2 a& cos 180° + b 2 sec 0°. 

229. m sin 90° - n cos 360° -f (m - ft) cos 180°. 



174 PROBLEMS FOR EXAMINATION. 

Prove the truth of the following equations : 

„_ sin x + tan x . , 

230. ■ = smajtanoj. 

cot X + CSC X 

The proof consists in reducing the equation to an identity. 
Thus, by clearing of fractions, both sides of the equation 
are sin x 4 tan x. By writing the work in reverse order, a 
direct proof of the given equation is obtained. 

231. tajig = a + sillfl; H 1 + 6eca? >. 

(1 -f- cos x) (1 + csc x) 

nr>r> „„„ „„„„„. sec x c °t x — csc x tan X 

232. csc x sec x = . 



233. tan x — cot x 



cos x — sin x 
1 — 2 cos 2 a; 



sin x cos x 

234. tan x + cot x = cot x sec 2 x. 

235. sin 2 x 4 tan 2 x = sec 2 x — cos 2 x. 

236. sec x — cos a; = sin x tan a;. 

___ sin x 4 cos # sec x + csc a? 
sin a; — cos x sec a; — csc x 

238. (sin a; 4 tan x) (cos £ 4 cot x) = (l + sin cc) (1 4 cos x). 

2 tan x 2 cot a? 



239. sin 2 x = 

240. tan 2 a; = 

241. sec 2 x = 



1 + tan 2 x 1 + cot 2 a? 

' 1 1 

1 — tan x 1 -f tan a; 

sec 2 a? sec x csc a? 



1 — tan 2 a; cot x — tan a; 

242. csc 2x — \ cot a; cos 2 x — \ (tan x 4- cot #). 

243. cot 2x — \ (cot a? — tan x) — \ sec x csc a; — tan x. 
cot 2 a; — 1 cot x — tan a? 



244. cos 2 a? 



cot 2 x + 1 cot a; 4 tan x 



GONIOMETRIC RELATIONS. 175 

245. sin (x -j- y) — (tan x + tan y) cos x cos y. 

^„_ , , . N sin 2 a; — sin 2 ?/ 

246. tan (a; + ?/) = — : — ^ 

sm x cos x — sm ?/ cos ?/ 

_._ / , N sec a; sec?/ 

247. sec (x + y) = — 

1 — tan x tan y 

248. sin (x-y) = tan x - tan y 
sin (a; + y) tan a; + tan y 

«„« , , tan x + tan?/ 

249. tan cc tan 2/ = ! -• 

cot a; + cot?/ 

250. tan ( + y) + tan (x - y) = /^"'^ ■ 

jl — tan x tan ?/ 

251. sin (#+?/) -f cos (x—y) = (sin#+cosa?)(sin?/-|-cos?/). 

oeo , , / N sin x — sin ?/ cos y — cos a 

252. tan 1 (a; — ?/) = - = - — * 

cos x + cos?/ sin y + sin a; 

253. tan (45° + x) = cos * + s |" * = 1 + taK * . 

cos a; — sm x 1 — tan a; 

tan x _ sin (a; -f y) + sin (a; — y) 
tan ?/ sin (x -\- y) — sin (a; — y) 

255. tan 2 \x-\-2 cot a; tan 1 a: — 1 = 0. 

256. tan 2 \x — 2 esc a; tan 1 x + 1 = 0. 

Explain how the general quadratic equation may be solved 
by the use of 255 or 256. 

__„ , , 1 + sin x — cos x 

257. tan i x — . 

1 + sin x + cos a; 

258. sin nx — 2 sin (n — l)x cos x — sin (w — 2)x. 

259. cos ?i# = 2 cos (w — l)a? cos x — cos (?i — 2)x. 

260. to M = tan (».-!>■ + tan x , 

1 — tan (?i — l)a; tan a: 



176 PROBLEMS FOR EXAMINATION. 

Solve the following equations : 

261. a tan 2 x + & = c sec a?. 

262. cos 2 x — sin 2 a? + tan 2 x = -J. 

263. sin (20° + «) cos (20° - x) = .2. 

264. tan x -+- cot x = 4 cot 2 #. 

265. cot # — tan x — A cos 2 x. 

266. tan 2 a? + tan 3 x = 3 tan a?. 

267. 1 — cos 2x = 3sin2x. 

268. sin # + cos x = V2. 

269. 16 (cot x — sec #) = tan x. 
2.10. tan x + cot jc = 2 sec 2 #. 

271. cot x tan 2 a — tan x cot 2 x = 2. 

272. sec 2 x + tan a; = 3. 

273. arc tan (JV + 1) = 3 arc tan(^T— 1). 



274. arc tan arc tan = J- 



N-l JV+1 



r2 



7T. 



ANSWERS. 



1. c tan A = 102.7. 2 1. a Vcot 2 A + cot 2 D = 109. 

2. a cot 12° 40' = 387.1. 22 d = 12 g 

3. cot a =2.4; 22° 37'. " Vcot 2 D - cot 2 .4 

4. 23 sec 38° = 29.19. 23. 6 2 sin2^1 = 6 2 - (a - c) 2 ; 

5. 880 sin 33° 3' =480. K22°32'E. 

6. c tan A = 125.1. 24. 4 J fir=918(a- & + c) ; 



7. c tan .4 + d = 18. 



340, 189. 



8. c (tan ^ + tan A) = 102. 25 " & 2 = (« + c) 2 - 4 A- 377. 

9. a (1 + tan ^ cot A) = 76.5. 26 " 13,7 > 30 - 8 ' 
10 atan^ _ 1(}() 27. 4.844,8.72. 

tan A 1 — tan ^1 28. 14.25, 8.75. 

1± a = 5Q0j 29. 154 tan 27° 10' + 5 = 84. 

tan ^4+ tan A 30 3034,52.94. 

12. a (tan C — tan Q) = 1.5. 31 ^go gQ, 

13. c (tan ^- tan J[) = 128.4. 32 12 g 54 o 52 ' 

14. -J* - = 600. 



cot A + cot A 1 33. 27ri?cos40 ° = ,2206. 



15 



c tan -iuLj />a 



86400 



tan A - tan ^ 34. 2° 18.7', .1615 m., 852.6 ft. 

16. V^ = 150. 35. tan x = - + n ; 77°. 

17 — ' f , =9. 36. i& V2 tan 43° 35' =1700. 

tan -o_i — tan .^L . 

18. (a - h) cot A = 3.1. 37. a\/^±-^=60. 

19. cot x = n : 51° 41 '. o k 

,mm. 3a ^F esc 33° 45' = 10. 

20. tana? = wj 47° 44'. 27 

177 



178 



39. 13.5 esc 58° 6' = 
100 



ANSWERS. 
15.9. 



40. 



21.8. 



cot 20° 15' + 2 cos 20° 15' 

41. sec A = 2A; A = 65° 23' ; K = 10.64. 

42. 1.5 d 2 sin 60° sin 2 \A = 8.798. 

43. Area = tayo triangles — sector = \:>r sin 60° ; 15. 

44. tt6 2 cos A = 201.2. 45. 7riTsec^C = 33.49. 

46. 36 7rF 2 = >S 3 sin 2 Ccsc4C; 73.43. 

47. racl base = 1.81 esc 54° 18'; V= 36.65. 

48. jh sec A = 2p ; K=p 2 cot ^4 = 25.57. 



49. tanJ. = |r; .4= 57° 31' 

50. 100°. 51. 50. 52. 48° 55' 
106.2" 



57. 2.143 7T cot 25° 43' =14 
90° 



53 



180 a 



= 6.185. 



58. 



54. 21. 55. 163° 34'. 

56. .3125 cZ 2 esc 2 36° = 12.15. 



iacsc 



2 71 + 1 

180° 
2/i + l 
59. 143.1 feet. 



a cos 



988. 
208. 



60. 27/- 2 tan6°40'sin 2 6°40' = 3.6. 

„ 9 m_ 360°, 180° 1ifiK 

61. 2— it esc tan = 116.o. 

n n ni 

62. A r 9 (16.96 7T + sin 70° 32 ' - sin 40°) = .4656. 

63. 441 cos 46° tan 20° = 157. 67. c sin {A 1 —A) sec A sec A v 

64. a esc (A 1 —A) cos ^1 sin ^ 68. c esc (^.-h^!) sin A sin ^4j. 

65. a esc (A-\-A±) cos ^4 cos A v 69. c esc (^4—^) cos ^4 sin A v 

66. a sin (C- Ci) see C sec Ci- 70. Apply (19) ; 890. 

71. Apply (20) ; 741.2. 

72. 18 sec 87° 30' cos 37° 30' sin 40° = 210. 

73. 50 esc 5° 55' cos 31° 12' sin 2o° 17' = 177.1. 
50 esc 5° 55' cos 25° 17' = 438.4. 

74. Apply (24) : X. 87° 49' E. 
Apply (19) ; 12 miles an hour. 



OBLIQUE TRIANGLES. 179 

75. 528 esc 37° sin 42° sec 29° sin 8° = 114.5. 

528 esc 37° sin 5° = 301. 

76. c esc C sin A sin B — 50 = 240. 

77. 775.3 esc 27° 35' sin 126° 43' sin 32° 57' = 730. 

78. 200 sec 2 26° 37' esc 10° 8' sin 10° 22' sin 20° 30' = 89.63. 

79. 180 esc 17° 15' sin 33° 45' sec 40° sin 11° = 84. 

80. Apply (27) ; 18.71 + 16.2 = 34.91. 

81. Apply (24) ; cot x = 1.76 esc 30° - cot 30°. 
29° 15' with the front of the train. 

82. 6 sin 45° esc 10° = 24.44. 85. 3.5. 

83. K 70° 44' E. ; 9.983. 86. Apply (30) ; 229,000. 

84. 10.32 ; 78° 14'. 87. Apply (29) ; 52. 

88. Apply (26) ; b = x cot A, I = x cot D, a = x cot B ; 60. 

89. Apply (19) ; 20.51 ; 23.63. 

90. Apply (16), (19), and (20) ; 

A c + 4 c + 1 K 

cos A = - — ' = — : c = 5. 

2 (c + 1) 2(c-l)' 

91. c cos \{B + A) sin \ (B — A) = 57. 

92. SnelPs ; 1034, 1465, 2047. 

93. 300 csc 68° 14' sin 120° 26' esc 17° 48' sin 70° = 856.2. 

94. 100 sin 88° 10' csc 8° 20' = 676. 95. Apply (25) ; 5.715. 
96. Apply (25) ; 331. 97. Apply (28) ; 1054. 

98. Find AB and BD and apply (20) ; 287.2. 

99. Find AB, then BC = .3537. 

100. sin ADB = tan 56° 2& cot 65°. 23' sin 56° 33'; 114. 

101. Apply (19) ; 410. 

102. Same as Problem 88 ; 100. 

103. Snell's ; 229, 61, 109. 

104. By (30) cos 2 ^=.85; .4=22° 47'. 

By (28) cot»=1.126 csc 45° 34'-cot45° 34'; z=59° 15'. 
By (19) y=9 csc 22° 47' sin 82° 2' cos 59° 15'=11.77. 



180 ANSWERS. 

105. i (12 csc A) (16 esc A) sin A = 96cscA = 100. 

106. Divide into a triangle and a parallelogram ; 46.82. 

_„ -, V + W ± V3 VIC n _, w_ 

107. ia — == = 9 sec. or 1 mm. 5< sec. 

v + vr — vw 

108. 70 sin i^ — A^sm\(B x + B^x 

csc 1 (^ + A 2 + A + -B 2 ) = 18.34. 

109. The pond is the incircle of the triangle formed by 
the three tangents ; 36.58. 

110. The pond is the excircle of the triangle formed by 
the three tangents ; 768.3. 

111. The hill is at the center of the circumcircle. 
60.8 csc 15° 25' tan 52° 40' = 300. 

112. r tan \B tan 1(7=1.589. 113. r tan 2 (±b-\A) =1.718. 

114. 4 r sin (45° -\A) sin (45° - 15) sin (45°.— \ C) =2.925. 

115. 3107. 116. 5057. 117. 2713. 
118. 2410 geographical miles. 119. 7 hours. 

120. lat. 37° 36'; long. 68° 38'. 

121. cos Z = - sec I sin d. 238° 42' ; 270°; 301° 18'; 180°. 
cos P = — tan I tan d. length = 2 P -h 15. 

14 hr. 50 min. 48 sec. ; 12 hr. ; 9 hr. 9 min. 12 sec. ; 24 hr. 

122. sec P = tan I cot d. 3 hr. 55 min. 28 sec. ; 6 ; 12 p.m. 

123. sin a = sin I sin d. 14° 49'. tan Z = sec I cot d. 251° 37'. 

124. tan 1 = tan d sec 31° 30'; 14° 46'. 

125. Let R be first position, S the second. 

2frS = 49°4P', PBS = 84° 2', ZBS = 50° 3V ; lat. = 47° 55'. 

126. P = 32° 30', ZS = 45° 40', PS = 116° 6' ; lat. = 6° 56'. 

127. 40° 6'. 128. .336 B 2 . 129. 2.443 B 2 . 
13Q /logd-cloga = _ 1121 131 log_e = 2W 

b log a — e log d &loga + c?logc 

132. Roots of x 2 log d — a; log b = log a — log c. Imaginary. 



GONIOMETRY. 



181 



133. x = log b (/log e ~ d log c) = - 21.93. 

log a log e — log b log c 



134. x 



" log 6 log a 
log a log 5 



1.108. 



135. Eoots of x 2 log a + x log 6 + log c = 0. Imaginary. 



136. 



log log c - log log a = 63()9 



137. tan x = 



log b 
6 sin 20° - sin 10 c 



2° 56' 



cos 10° - .6 cos 20 c 

138. Eeduce to sines and cosines and apply (69), 

sin (2 x - 10°) = - 5 sin 10° ; 125° V. 

139. Apply the value of tan (x ± y) and reduce 

cot 2 x = 1 + 2 tan 2 10° ; 44° 8'. 

140. Apply (65) and (75) ; cot x= -(2 + cot 10°); 172° 34'. 

141. cos (2 x - 10°) = 1.2 - cos 10° ; 43° 47'. 

142. By (13) ; 32 cos 2 x = 16 + 16 cos 2 x. 
16cos 2 2a + 16cos2a; + 3 = 0; a = 52°14'. 

143. Apply (69); 29° 33'. 144. 4cosa= vTf-1; 38° 40'. 

145. sin (2 a?- 70°) = .6 esc 70° ; 105° 10'. 

146. tan x= .4142 ; 22° 30'. 147. tan x = tan a tan b tan c. 

148. 160 tan A = 70 tan (.4 + 21°) = 168. 

149. a tan A = (a — 5) tan 3 ^L = 59.2. 

150. 97° 37' with the road; 11.44 miles. 

151. 360 cot A = 360 cot (A - 2° 7') = 4994. 

152. i r 2 (0 - sin 0) = 13.714 and 4.252 ; 9.462. 
Trr 2 - (13.714 + 4.252) = 1239. 

153. Two segments ; 0, = 153° 44', 2 = 54° 33'; .1 A. 



182 FORMULAE FOR RIGHT TRIANGLES. 




esc A 



tan A 



sin A 



cos A 



PLANE RIGHT TRIANGLES. 

Any side = adjacent side times adjacent function. 

Any function of A = adjacent side divided by the next side. 

Any function of C = the corresponding cofunction of A. 

SPHERICAL RIGHT TRIANGLES. 



Any function is equal to 
the product of its adjacent 
functions. 

Any function is equal to 
the product of an adjacent 
function and the reciprocal 
of the next function. 




tan a — cos O 



A TABLE OF 

FOUR-PLACE MANTISSA OF LOGARITHMS 
OF NUMBERS 

FROM 1 TO 1000 



A TABLE OF 

THE LOGARITHMS OF THE TRIGONOMETRIC 
FUNCTIONS OF ANGLES 

FROM 0° TO 90° FOR EVERY 10' 



A TABLE OF FORMULAE 



A TABLE OF VALUES OF CONSTANTS 



Vlll 



LOGAKITHMS OF NUMBERS. 



N 



2 



8 



10 

11 
12 
13 
14 

15 

16 
17 

18 
19 

20 

21 
22 
23 
24 

25 

26 

27 
28 
29 

30 

31 
32 
33 

34 

35 

36 

37 
38 
39 

40 

41 
42 
43 
44 

45 
46 

47 
48 

49 

50 

51 

52 
53 
54 



.0000 
414 
792 

.1139 
461 

.1761 

.2041 

304 

553 

788 

.3010 
222 
424 
617 
802 

.3979 

.4150 

314 

472 

624 

.4771 
914 

.5051 
185 
315 

.5441 
563 
682 
798 
911 

.6021 
128 
232 
335 

435 

.6532 
628 
721 
812 
902 

.6990 

.7076 

160 

243 

324 



.0043 
453 
828 

.1173 
492 

.1790 

2068 

330 

577 

810 

.3032 
243 
444 
636 
820 

.3997 

.4166 

330 

487 

639 

.4786 
928 

.5065 
198 
328 

.5453 
575 
694 
809 
922 

.6031 
138 
243 
345 
444 

.6542 
637 
730 
821 
911 

.6998 

.7084 

168 

251 

332 



.0086 
492 
864 

.1206 
523 

.1818 

.2095 

355 

601 

833 

.3054 
263 
464 
655 
838 

.4014 
183 
346 
502 
654 

.4800 
942 

.5079 
211 
340 

.5465 
587 
705 
821 
933 

.6042 
149 
253 
355 

454 

.6551 
646 
739 
830 
920 

.7007 
093 
177 
259 
340 



.0128 
531 
899 

.1239 
553 

.1847 

.2122 

380 

625 

856 

.3075 
284 
483 
674 
856 

.4031 
200 
362 
518 
669 

.4814 
955 

.5092 
224 
353 

.5478 
599 
717 
832 
944 

.6053 
160 
263 
365 
464 

.6561 
656 
749 
839 
928 

.7016 
101 
185 
267 
348 



.0170 
569 
934 

.1271 

584 

.1875 

.2148 

405 

648 

878 

.3096 
304 
502 
692 

874 

.4048 
216 
378 
533 
683 

.4829 
969 

.5105 
237 
366 

.5490 
611 
729 
843 
955 

.6064 
170 
274 
375 
474 

.6571 
665 

758 
848 
937 

.7024 
110 
193 

275 
356 



.0212 
607 
969 

.1303 
614 

.1903 

.2175 

430 

672 

900 

.3118 
324 
522 
711 
892 

.4065 
232 
393 
548 
698 

.4843 
983 

.5119 
250 
378 

.5502 
623 
740 
855 
966 

.6075 
180 
284 
385 
484 

.6580 
675 
767 
857 
946 

.7033 
118 
202 
284 
364 



.0253 .0294 
645 682 

.1004 .1038 
335 367 
644 673 

.1931 .1959 

.2201 .2227 

455 480 

695 718 

923 945 

.3139 .3160 
345 365 
541 560 
729 747 
909 927 

.4082 .4099 
249 265 
409 425 
564 579 
713 728 

.4857 .4871 
997 .5011 

.5132 145 
263 276 
391 403 

.5514 .5527 
635' 647 
752 763 
866 877 
977 988 



.0334 .0374 
719 755 

.1072 .1106 
399 430 
703 732 

.1987 .2014 

.2253 279 

504 529 

742 765 

967 989 

.3181 .3201 
385 404 
579 598 
766 784 
945 962 



.6085 
191 
294 
395 
493 

.6590 
684 
776 
866 
955 

,7042 
126 
210 
292 

372 



.6096 
201 
304 
405 
503 

.6599 
693 

785 
875 
964 

.7050 
135 
218 
300 
380 



.4116 
281 
440 
594 
742 

.4886 

.5024 

159 

289 

416 

.5539 
658 

775 
888 
999 

.6107 
212 
314 
415 
513 

.6609 
702 
794 
884 
972 

.7059 
143 

226 
308 

388 



.4133 
298 
456 
609 

757 

.4900 

.5038 

172 

302 

428 

.5551 
670 

786 

899 

.6010 

.6117 

222 
325 
425 

522 

.6618 
712 
803 
893 
981 

.7067 
152 
235 
316 
396 



LOGARITHMS OF NUMBERS. 



IX 



N 





1 


2 


3 


4 


5 


6 


7 


8 9 


55 


.7404 


.7412 


.7419 


.7427 


.7435 


.7443 


.7451 


.7459 


.7466 .7474 


56 


482 


490 


497 


505 


513 


520 


528 


536 


543 551 


57 


559 


566 


574 


582 


589 


597 


604 


612 


619 627 


58 


634 


642 


649 


657 


664 


672 


679 


686 


694 701 


59 


709 


716 


723 


731 


738 


745 


752 


760 


767 774 


60 


.7782 


.7789 


.7796 


.7803 


.7810 


.7818 


.7825 


.7832 


.7839 .7846 


61 


853 


860 


868 


875 


882 


889 


896 


903 


910 917 


62 


924 


931 


938 


945 


952 


959 


966 


973 


980 987 


63 


993 


.8000 


.8007 


.8014 


.8021 


.8028 


.8035 


.8041 


.8048 .8055 


64 


.8062 


069 


075 


082 


089 


096 


102 


109 


116 122 


65 


.8129 


.8136 


.8142 


.8149 


.8156 


.8162 


.8169 


.8176 


.8182 .8189 


66 


195 


202 


209 


215 


222 


228 


235 


241 


248 254 


67 


261 


267 


274 


280 


287 


293 


299 


306 


312 319 


68 


325 


331 


338 


344 


351 


357 


363 


370 


376 382 


69 


388 


395 


401 


407 


414 


420 


426 


432 


439 445 


70 


.8451 


.8457 


.8463 


.8470 


.8476 


.8482 


.8488 


.8494 


.8500 .8506 


71 


513 


519 


525 


531 


537 


543 


549 


555 


561 567 


72 


573 


579 


585 


591 


597 


603 


609 


615 


621 627 


73 


633 


639 


645 


651 


657 


663 


669 


675 


681 686 


74 


692 


698 


704 


710 


716 


722 


727 


733 


739 745 


75 


.8751 


.8756 


.8762 


.8768 


.8774 


.8779 


.8785 


.8791 


.8797 .8802 


76 


808 


814 


820 


825 


831 


837 


842 


848 


854 859 


77 


865 


871 


876 


882 


887 


893 


899 


904 


910 915 


78 


921 


927 


932 


938 


943 


949 


954 


960 


965 971 


79 


976 


982 


987 


993 


998 


.9004 


.9009 


.9015 


.9020 .9025 


80 


.9031 


.9036 


.9042 


.9047 


.9053 


.9058 


.9063 


.9069 


.9074 .9079 


81 


085 


090 


096 


101 


106 


112 


117 


122 


128 133 


82 


138 


143 


149 


154 


159 


165 


170 


175 


180 186 


83 


191 


196 


201 


206 


212 


217 


222 


227 


232 238 


84 


243 


248 


253 


258 


263 


269 


274 


279 


284 289 


85 


.9294 


.9299 


.9304 


.9309 


.9315 


.9320 


.9325 


.9330 


.9335 .9340 


86 


345 


350 


355 


360 


365 


370 


375 


380 


385 390 


87 


395 


400 


405 


410 


415 


420 


425 


430 


435 440 


88 


445 


450 


455 


460 


465 


469 


474 


479 


484 489 


89 


494 


499 


504 


509 


513 


518 


523 


528 


533 538 


90 


.9542 


.9547 


.9552 


.9557 


.9562 


.9566 


.9571 


.9576 


.9581 .9586 


91 


590 


595 


600 


605 


609 


614 


619 


624 


628 633 


92 


638 


643 


647 


652 


657 


661 


666 


671 


675 680 


93 


685 


689 


694 


699 


703 


708 


713 


717 


722 727 


94 


731 


736 


741 


745 


750 


754 


759 


763 


768 773 


95 


.9777 


.9782 


.9786 


.9791 


.9795 


.9800 


.9805 


.9809 


.9814 .9818 


96 


823 


827 


832 


836 


841 


845 


850 


854 


859 863 


97 


868 


872 


877 


881 


886 


890 


894 


899 


903 908 


98 


912 


917 


921 


926 


930 


934 


939 


943 


948 952 


99 


956 


961 


965 


969 


974 


978 


983 


987 


991 996 



LOGARITHMS OF FUNCTIONS. 



X 


L sin* 


L tan* 


log sec x 


log CSC X log COt X 


/.cos* 




0° 


Neg. Inf. 


Neg. Inf. 


0.000 


Infinite Infinite 10.000 


90° 


10' 


7^ 4637 


7. 4637 





2. 5363 2. 5363 





50* 


20' 


7648 


7648 





2352 2352 





40' 


30' 


9408 


9409 





0592 0591 





30' 


40' 


8. 0658 


8. 0658 





1. 9342 1. 9342 





20' 


50' 


1627 


1627 





8373 8373 





10' 


1° 


8. 2419 


8. 2419 


0.000 1 


1. 7581 1. 7581 


9.999 9 


89° 


10' 


3088 


3089 


1 


6912 6911 


9 


50' 


20' 


3668 


3669 


1 


6332 6331 


9 


40' 


30' 


4179 


4181 


1 


5821 5819 


9 


30' 


40' 


4637 


4638 


2 


5363 5362 


8 


20' 


50' 


5050 


5053 


2 


4950 4947 


8 


10' 


2° 


8. 5428 


8. 5431 


0.000 3 


1. 4572 1. 4569 


9.999 7 


88° 


10' 


5776 


5779 


3 


4224 4221 


7 


50' 


20' 


6097 


6101 


4 


3903 3899 


6 


40' 


30' 


6397 


6401 


4 


3603 3599 


6 


30' 


40' 


6677 


6682 


5 


3323 3318 


5 


20' 


50' 


6940 


6945 


5 


3060 3055 


5 


10' 


3° 


8. 7188 


8. 7194 


0.000 6 


1. 2812 1. 2806 


9.999 4 


87° 


10' 


7423 


7429 


7 


2577 2571 


3 


50' 


20' 


7645 


7652 


7 


2355 2348 


3 


40' 


30' 


7857 


7865 


8 


2143 2135 


2 


30' 


40' 


8059 


8067 


9 


1941 1933 


1 


20' 


50' 


8251 


8261 


0.001 


1749 1739 





10' 


4° 


8. 8436 


8. 8446 


0.001 1 


1. 1564 1. 1554 


9.998 9 


86° 


10' 


8613 


8624 


1 


1387 1376 


9 


50' 


20' 


8783 


8795 


2 


1217 1205 


8 


40' 


30' 


8946 


8960 


3 


1054 1040 


7 


30' 


40' 


8. 9104 


8. 9118 


4 


1. 0896 1. 0882 


6 


20' 


50' 


9256 


9272 


5 


0744 072S 


5 


10' 


5° 


8. 9403 


8. 9420 


0.001 7 


1. 0597 1. 0580 


9.998 3 


85° 


10' 


9545 


9563 


8 


0455 0437 


2 


50' 


20' 


9682 


9701 


9 


0318 0299 


1 


40' 


30' 


9S16 


9836 


0.002 


0184 0164 





30' 


40' 


9945 


9966 


1 


0055 0034 


9.997 9 


20' 


50' 


9. 0070 


9. 0093 


3 


0. 9930 0. 9907 


7 


10' 


6° 


9. 0192 


9. 0216 


0.002 4 


0. 9808 0. 9784 


9.997 6 


84° 


10' 


0311 


0336 


5 


9689 9664 


5 


50' 


20' 


0426 


0453 


7 


9574 9547 


3 


40' 


30' 


0539 


0567 


8 


9461 9433 


2 


30' 


40' 


0648 


0678 


9 


9352 9322 


1 


20' 


50' 


0755 


0786 


0.003 1 


9245 9214 


9996 9 


10' 


7° 


0859 


9. 0891 


0.003 2 


9141 9109 


9.996 8 


83° 




L cos/ 


/.cot/ 


log CSC/ 


log sec/ log tan/ 


L sin/ 


/ 



LOGARITHMS OF FUNCTIONS. 



XI 



X 


L sin* 


L tan* 


iog sec * 


lOg CSC * 


L cot x L cos x 




7° 


9.0 859 


9.0 891 


0.00 32 


0.9 141 


0.9 109 9.99 68 


83° 


10' 


961 


995 


34 


039 


005 66 


50' 


20' 


9.1 060 


9.1 096 


36 


0.8 940 


0.8 904 64 


40' 


30' 


157 


194 


37 


843 


806 63 


30' 


40' 


252 


291 


39 


748 


709 61 


20' 


50' 


345 


385 


0.00 41 


665 


615 59 


10' 


8° 


9.1 436 


9.1 478 


0.00 42 


0.8 564 


0.8 522 9.99 58 


82° 


10' 


525 


569 


44 


475 


431 56 


50' 


20' 


612 


658 


46 


388 


342 54 


40' 


30' 


697 


745 


48 


303 


255 52 


30' 


40' 


781 


831 


50 


219 


169 50 


20' 


50' 


863 


915 


52 


137 


085 48 


10' 


9° 


9.1 943 


9.1 997 


0.00 54 


0.8 057 


0.8 003 9.99 46 


81° 


10' 


9.2 022 


9.2 078 


56 


0.7 978 


0.7 922 44 


50' 


20' 


100 


158 


58 . 


900 


842 42 


40' 


30' 


176 


236 


60 


824 


764 40 


30' 


40' 


251 


313 


62 


749 


687 38 


20' 


50' 


324 


389 


64 


676 


611 36 


10' 


10° 


9.2 397 


9.2 463 


0.00 66 


0.7 603 


0.7 537 9.99 34 


80° 


10' 


468 


536 


69 


532 


464 31 


50' 


20' 


538 


609 


71 


462 


391 29 


40' 


30' 


606 


680 


73 


394 


320 27 


30' 


40' 


674 


750 


76 


326 


250 24 


20' 


50' 


740 


819 


78 


260 


181 22 


10' 


11° 


9.2 806 


9.2 887 


0.00 81 


0.7 194 


0.7 113 9.99 19 


79° 


10' 


870 


953 


83 


130 


047 17 


50' 


20' 


934 


9.3 020 


86 


066 


0.6 980 14 


40' 


30' 


997 


085 


88 


003 


915 12 


30' 


40' 


9.3 058 


149 


91 


0.6 942 


851 09 


20' 


50' 


119 


212 


93 


S81 


788 07 


10' 


12° 


9.3 179 


9.3 275 


0.00 96 


0.6 821 


0.6 725 9.99 04 


78 


10' 


238 


336 


99 


762 


664 01 


50' 


20' 


296 


397 


0.01 01 


704 


603 9.98 99 


40' 


30' 


353 


458 


04 


647 


542 96 


30' 


40' 


410 


517 


07 


590 


483 93 


20' 


50' 


466 


576 


10 


534 


424 90 


10' 


13° 


9.3 521 


9.3 634 


0.01 13 


0.6 479 


0.6 366 9.98 87 


77° 


10' 


575 


691 


16 


425 


309 84 


50' 


20' 


629 


748 


19 


371 


252 81 


40' 


30' 


682 


804 


22 


318 


196 78 


30' 


40' 


734 


859 


25 


266 


141 75 


20' 


50' 


786 


914 


28 


214 


086 72 


10' 


14° 


9.3 837 


9.3 968 


0.01 31 


0.6 163 


0.6 032 9.98 69 


76° 




L cos/ 


L cot/ 


log CSC/ 


log sec/ 


log tan/ L sin/ 


y 



Xll 



LOGAKITHMS OF FUNCTIONS. 



X 


L sin* 


L tan* 


log sec x 


log CSC X 


log cot X 


Lcosx 




14° 


9.3 837 


9.3 968 


0.01 31 


0.6 163 


0.6 032 


9.98 69 


76° 


10' 


887 


9.4 021 


34 


113 


0.5 979 


66 


50' 


20' 


937 


074 


37 


063 


926 


63 


40' 


30' 


986 


127 


41 


014 


873 


59 


30' 


40' 


9.4 035 


178 


44 


0.5 965 


822 


56 


20' 


50' 


083 


230 


47 


917 


770 


53 


10' 


15° 


9.4 130 


9.4 281 


0.01 51 


0.5 870 


0.5 719 


9.98 49 


75° 


10' 


177 


331 


54 


823 


669 


46 


50' 


20' 


223 


381 


57 


777 


619 


43 


40' 


30' 


269 


430 


61 


731 


570 


39 


30' 


40' 


314 


479 


64 


686 


521 


36 


20' 


50' 


359 


527 


68 


641 


473 


32 


10' 


16° 


9.4 403 


9.4 575 


0.01 72 


0.5 597 


0.5 425 


9.98 28 


74° 


10' 


447 


622 


75 


553 


378 


25 


50' 


20' 


491 


669 


79 


509 


331 


21 


40' 


30' 


533 


716 


83 


467 


284 


17 


30' 


40' 


576 


762 


86 


424 


238 


14 


20' 


50' 


618 


808 


90 


382 


192 


10 


10' 


17° 


9.4 659 


9.4 853 


0.01 94 


0.5 341 


0.5 147 


9.98 06 


73° 


10' 


700 


898 


98 


300 


102 


02 


50' 


20' 


741 


943 


0.02 02 


259 


057 


9.97 98 


40' 


30' 


781 


987 


06 


219 


013 


94 


30' 


40' 


821 


9.5 031 


10 


179 


0.4 969 


90 


20' 


50' 


861 


075 


14 


139 


925 


86 


10' 


18° 


9.4 900 


9.5 118 


0.02 18 


0.5 100 


0.4 882 


9.97 82 


72° 


10' 


939 


161 


22 


061 


839 


78 


50' 


20' 


977 


203 


26 


023 


797 


74 


40' 


30' 


9.5 015 


245 


30 


0.4 985 


755 


70 


30' 


40' 


052 


287 


35 


948 


713 


65 


20' 


50' 


090 


329 


39 


910 


671 


61 


10' 


19° 


9.5 126 


9.5 370 


0.02 43 


0.4 874 


0.4 630 


9.97 57 


71° 


10' 


163 


411 


48 


837 


589 


52 


50' 


20' 


199 


451 


52 


801 


549 


48 


40' 


30' 


235 


491 


57 


765 


509 


43 


30' 


40' 


270 


531 


61 


730 


469 


39 


20' 


50' 


306 


571 


66 


694 


429 


34 


10' 


20° 


9.5 341 


9.5 611 


0.02 70 


0.4 659 


0.4 389 


9.97 30 


70° 


10' 


375 


650 


75 


625 


350 


25 


50' 


20' 


409 


689 


79 


591 


311 


21 


40' 


30' 


443 


727 


84 


557 


273 


16 


30' 


40' 


477 


766 


89 


523 


234 


11 


20' 


50' 


510 


804 


94 


490 


196 


06 


10' 


21° 


9.5 543 


9.5 842 


0.02 98 


0.4 457 


0.4 158 


9.97 02 


69° 




L cosy 


L cot/ log cscy 


log secy logtany 


L sin/ 


Y 



LOGARITHMS OF FUNCTIONS. 



Xlll 



X 


Asm* 


L tanx 


log sec x 


log csc * 


log cot X 


L cos* 




21° 


9.5 543 


9.5 842 


0.02 98 


0.4 457 


0.4 158 


9.97 02 


69° 


10' 


576 


879 


0.03 03 


424 


121 


9.96 97 


50' 


20' 


609 


917 


08 


391 


083 


92 


40' 


30' 


641 


954 


13 


359 


046 


87 


30' 


40' 


673 


991 


18 


327 


009 


82 


20' 


50' 


704 


9.6 028 


23 


296 


0.3 972 


77 


10' 


22° 


9.5 736 


9.6 064 


0.03 28 


0.4 264 


0.3 936 


9.96 72 


68° 


10' 


767 


100 


33 


233 


900 


67 


50' 


20' 


798 


136 


39 


202 


864 


61 


40' 


30' 


828 


172 


44 


172 


828 


56 


30' 


40' 


859 


208 


49 


141 


792 


51 


20' 


50' 


889 


243 


54 


111 


757 


46 


10' 


23° 


9.5 919 


9.6 279 


0.03 60 


0.4 081 


0.3 721 


9.96 40 


67° 


10' 


948 


314 


65 


052 


686 


35 


50' 


20' 


978 


348 


71 


022 


652 


29 


40' 


30' 


9.6 007 


383 


76 


0.3 993 


617 


24 


30' 


40' 


036 


417 


82 


964 


583 


18 


20' 


50' 


065 


452 


87 


935 


548 


13 


10' 


24° 


9.6 093 


9.6 486 


003 93 


0.3 907 


0.3 514 


9.96 07 


66° 


10' 


121 


520 


98 


879 


480 


02 


50' 


20' 


149 


553 


0.04 04 


851 


447 


9.95 96 


40' 


30' 


177 


587 


10 


823 


413 


90 


30' 


40' 


205 


620 


16 


795 


380 


84 


20' 


50' 


232 


654 


21 


768 


346 


79 


10' 


25° 


9.6 259 


9.6 687 


0.04 27 


0.3 741 


0.3 313 


9.95 73 


65° 


10' 


286 


720 


33 


714 


280 


67 


50' 


20' 


313 


752 


39 


687 


248 


61 


40' 


30' 


340 


785 


45 


660 


215 


55 


30' 


40' 


366 


817 


51 


634 


183 


49 


20' 


50' 


392 


850 


57 


608 


150 


43 


10' 


26° 


9.6 418 


9.6 882 


0.04 63 


0.3 582 


0.3 118 


9.95 37 


64° 


10' 


444 


914 


70 


556 


086 


30 


50' 


20' 


470 


946 


76 


530 


054 


24 


40' 


30' 


495 


977 


82 


505 


023 


18 


30' 


40' 


521 


9.7 009 


88 


479 


0.2 991 


12 


20' 


50' 


546 


040 


95 


454 


960 


05 


10' 


27° 


9.6 570 


9.7 072 


0.05 01 


0.3 430 


0.2 928 


9.94 99 


63° 


10' 


595 


103 


08 


405 


897 


92 


50' 


20' 


620 


134 


14 


380 


866 


86 


40' 


30' 


644 


165 


21 


356 


835 


79 


30' 


40' 


668 


196 


27 


332 


804 


73 


20' 


50' 


692 


226 


34 


308 


774 


66 


10' 


28° 


9.6 716 


9.7 257 


0.05 41 


0.3 284 


0.2 743 


9.94 59 


62° 




L cos/ 


Lcoty 


log CSC/ 


log sec/ log tan/ 


L sin/ 


' 



XIV 



LOGARITHMS OF FUNCTIONS. 



X 


L sin x 


L tan* 


log sec* 


log csc x 


log cot* 


L cos* 




28 


9.6 716 


9.7 257 


0.05 41 


0.3 2S4 


0.2 743 


9.94 59 


62 : 


10' 


740 


287 


47 


260 


713 


53 


50' 


20' 


763 


317 


54 


237 


6S3 


46 


40' 


30' 


787 


348 


61 


213 


652 


39 


30' 


40' 


810 


37S 


68 


190 


622 


32 


20' 


50' 


833 


40S 


75 


167 


592 


25 


10' 


29 : 


9.6 S56 


9.7 43S 


0.05 S2 


0.3 144 


0.2 562 


9.94 IS 


61° 


10' 


S7S 


467 


S9 


122 


533 


11 


50' 


20' 


901 


497 


96 


099 


503 


04 


40' 


30' 


923 


526 


0.06 03 


077 


474 9.93 97 


30' 


40' 


946 


556 


10 


054 


444 


90 


20' 


50' 


96S 


585 


17 


032 


415 


S3 


10' 


30 : 


9.6 990 


9.7 614 


0.06 25 


0.3 010 


0.2 3S6 


9.93 75 


60^ 


10' 


9.7 012 


644 


32 


0.2 9SS 


356 


68 


50' 


20' 


033 


673 


39 


967 


327 


61 


40' 


30' 


055 


701 


47 


945 


399 


53 


30' 


40' 


076 


730 


54 


924 


270 


46 


20' 


50' 


097 


759 


62 


903 


241 


3S 


10' 


31° 


9.7 US 


9.7 788 


0.06 69 


0.2 SS2 


0.2 212 


9.93 31 


59 


10' 


139 


S16 


77 


861 


184 


23 


50' 


20' 


160 


S45 


85 


840 


155 


15 


40' 


30' 


1S1 


873 


92 


819 


127 


OS 


30' 


40' 


201 


902 


0.07 00 


799 


09S 


00 


20' 


50' 


222 


930 


OS 


778 


070 


9.92 92 


10' 


32° 


9.7 242 


9.7 958 


0.07 16 


0.2 75 S 


0.2 042 


9.92 84 


58 : 


10' 


262 


9S6 


24 


73S 


014 


76 


50' 


20' 


282 


9.8 014 


32 


71S 


0.1 986 


. 68 


40' 


30' 


302 


042 


40 


698 


958 


60 


30' 


40' 


322 


070 


4S 


678 


930 


52 


20' 


50' 


342 


097 


56 


65 S 


903 


44 


10' 


33 : 


9.7 361 


9.8 125 


0.07 64 


0.2 639 


0.1 S75 


9.92 36 


57 : 


10' 


380 


153 


72 


620 


847 


28 


50' 


20' 


400 


180 


81 


600 


820 


19 


40' 


30' 


419 


203 


89 


581 


792 


11 


30' 


40' 


438 


235 


97 


562 


765 


03 


20' 


50' 


457 


263 


0.08 06 


543 


737 


9.91 94 


10' 


34 c 


9.7 476 


9.8 290 


0.08 14 


0.2 524 


0.1 710 


9.91 86 


56 : 


10' 


494 


317 


23 


506 


6S3 


77 


50' 


20' 


513 


344 


31 


4S7 


656 


69 


40' 


30' 


531 


371 


40 


469 


629 


60 


30' 


40' 


550 


398 


49 


450 


602 


51 


20' 


50' 


568 


425 


58 


432 


575 


42 


10' 


35 : 


9.7 586 


9.8 452 


0.0S 66 


0.2 414 


0.1 54S 


9.91 34 


55 




L cos/ 


L cot/ 


log CSC/ 


log sec/ 


log tan/ 


L sin/ 


/ 



LOGARITHMS OF FUNCTIONS. 



XV 



X 


L sin* 


L tan* 


log sec x 


log CSC X 


log cot X 


Lcosx 




35° 


9.7 586 


9.8 452 


0.08 66 


0.2 414 


0.1 548 


9.91 34 


55° 


10' 


604 


479 


75 


396 


521 


25 


50' 


20' 


622 


506 


84 


378 


494 


16 


40' 


30' 


640 


533 


93 


360 


467 


07 


30' 


40' 


657 


559 


0.09 02 


343 


441 


9.90 98 


20' 


50' 


675 


586 


11 


325 


414 


89 


10' 


36° 


9.7 692 


9.8 613 


0.09 20 


0.2 308 


0.1387 


9.90 80 


54° 


10' 


710 


639 


30 


290 


361 


70 


50' 


20' 


727 


666 


39 


273 


334 


61 


40' 


30' 


744 


692 


48 


256 


308 


52 


30' 


40' 


761 


718 


58 


239 


282 


42 


20' 


50' 


778 


745 


67 


222 


255 


33 


10' 


37° 


9,7 795 


9.8 771 


0.09 77 


0.2 205 


0.1 229 


9.90 23 


53° 


10' 


811 


797 


86 


189 


203 


14 


50' 


20' 


828 


824 


96 


172 


176 


04 


40' 


30' 


844 


850 


0.10 05 


156 


150 


9.89 95 


30' 


40' 


861 


876 


15 


139 


124 


85 


20' 


50' 


877 


902 


25 


123 


098 


75 


10' 


38° 


9.7 893 


9.8 928 


0.10 35 


0.2 107 


0.1 072 


9.89 65 


52° 


10' 


910 


954 


45 


090 


046 


55 


50' 


20' 


926 


980 


55 


074 


020 


45 


40' 


30' 


941 


9.9 006 


65 


059 


0.0 994 


35 


30' 


40' 


957 


032 


75 


043 


968 


25 


20' 


50' 


973 


058 


85 


027 


942 


15 


10' 


39° 


9.7 989 


9.9 084 


0.10 95 


0.2 011 


0.0 916 


9.89 05 


51° 


10' 


9.8 004 


110 


0.11 05 


0.1 996 


890 


9.88 95 


50' 


20' 


020 


135 


16 


980 


865 


84 


40' 


30' 


035 


161 


26 


965 


839 


74 


30' 


40' 


050 


187 


36 


950 


813 


64 


20' 


50' 


066 


212 


47 


934 


788 


53 


10' 


40° 


9.8 081 


9.9 238 


0.11 57 


0.1 919 


0.0 762 


9.88 43 


50° 


10' 


096 


264 


68 


904 


736 


32 


50' 


20' 


111 


289 


79 


889 


711 


21 


40' 


30' 


125 


315 


90 


875 


685 


10 


30' 


40' 


140 


341 


0.12 00 


860 


659 


00 


20' 


50' 


155 


366 


11 


845 


634 


9.87 89 


10' 


41° 


9.8 169 


9.9 392 


0.12 22 


0.1 831 


0.0 608 


9.87 78 


49° 


10' 


184 


417 


33 


816 


583 


67 


50' 


20' 


198 


443 


44 


802 


557 


56 


40' 


30' 


213 


468 


55 


787 


532 


45 


30' 


40' 


227 


494 


67 


773 


506 


33 


20' 


50' 


241 


519 


7S 


759 


481 


22 


10' 


42° 


9.8 255 


9.9 544 


0.12 89 


0.1 745 


0.0 456 


9.87 11 


48° 




/.cos/ 


L cot/ 


log CSC/ 


log sec/ log tan/ 


L sin/ 


y 



XVI 



LOGARITHMS OF FUNCTIONS. 



X 


L sinx 


Atari* 


log sec x 


log csc x 


log cot X 


L cos* 


48° 


42 Q 


9.8 255 


9.9 544 


0.12 89 


0.1 745 


0.0 456 


9.87 11 


10' 


269 


570 


0.13 01 


731 


430 


9.86 99 


50' 


20' 


283 


595 


12 


717 


405 


88 


40' 


30' 


297 


621 


24 


703 


379 


76 


30' 


40' 


311 


646 


35 


689 


354 


65 


20' 


50' 


324 


671 


47 


676 


329 


53 


10' 


43° 


9.8 338 


9.9 697 


0.13 59 


0.1 662 


0.0 303 


9.86 41 


47° 


10' 


351 


722 


71 


649 


278 


29 


50' 


20' 


365 


747 


82 


635 


253 


18 


40' 


30' 


378 


772 


94 


622 


228 


06 


30' 


40' 


391 


798 


0.14 06 


609 


202 


9.85 94 


20' 


50' 


405 


823 


18 


595 


177 


82 


10' 


44° 


9.8 418 


9.9 848 


0.14 31 


0.1 582 


0.0 152 


9.85 69 


46° 


10' 


431 


874 


43 


569 


126 


57 


50' 


20' 


444 


899 


55 


556 


101 


45 


40' 


30' 


457 


9.9 924 


68 


543 


0.0 076 


32 


30' 


40' 


469 


949 


80 


531 


051 


20 


20' 


50' 


482 


975 


93 


518 


025 


07 


10' 


45° 


9.8 495 


10.0 000 


0.15 05 


0.1 505 


0.0 000 


9.84 95 


45° 




L cos/ 


L cot/ 


log CSC/ 


log sec/ log tan/ 


L sin/ 


/ 



FORMULJE. 

a csc^4=& csc B= c csc C. 

«2 = &2 + c 2_2&cCOSA 

cot _B=c^^ -cot A 
b 

,_-yl(s-«)0-ft)(S-C) 



cot A J. = : cot \B= — 

r r 



Circle: circum=2 7rr ; area=7rr 2 . 
sector =%rx arc; seg=ir 2 (0— sin 5). 
Sphere: sur=4 7ri? 2 ; vol=f irB z . 
zone=2TrBh; seg=i7r/i 2 (3 B-h). 
Natural log N= 2. 30259 log 10 X. 
Pendulum: ir 2 l = gt. 
Falling bodies: 2s=gt 2 =vt=v 2 -t-g. 



NUMBER. LOG. 

radian 57.29578° 1.75812 

7T 180° 2.25527 

7T 10800' 4.03342 

7T 648000" 5.81158 

7T 3.14159 0.49715 

1-7T 0.31831 9.50285 

7T 2 9.86960 0.99430 

V^ 1.77245 0.24S57 

1 mile 52S0 ft. 3.72263 

1 acre 43560 sq. ft. 4.63909 

e 2.71S28 0.43429 

1-3/ 2.30259 0.36221 



el. light 186330 


5.27028 


Z(40°) 39.0986 


1.59216 


#(40°) 32.1573 


1.50728 



JUL ' i°30 



